452 Chapter 15

Solutions

CHAPTER 15

What Yoursquoll LearnYou will describe and cate-gorize solutions

You will calculate concen-trations of solutions

You will analyze the colliga-tive properties of solutions

You will compare and con-trast heterogeneousmixtures

Why Itrsquos ImportantThe air you breathe the fluidsin your body and some of thefoods you ingest are solu-tions Because solutions areso common learning abouttheir behavior is fundamentalto understanding chemistry

Though it isnrsquot apparent thereare at least three different solu-tions in this photo the air thelake in the foreground and thesteel used in the construction ofthe buildings are all solutions

Visit the Chemistry Web site atchemistrymccom to find linksabout solutions

151 What are solutions 453

DISCOVERY LAB

Materials

balance50-mL graduated cylinder100-mL beaker (2)stirring rodammonium chloride (NH4Cl)calcium chloride (CaCl2)water

Solution Formation

The intermolecular forces among dissolving particles and theattractive forces between solute and solvent particles result in

an overall energy change Can this change be observed

Safety PrecautionsDispose of solutions by flushing them down a drainwith excess water

Procedure

1 Measure 10 g of ammonium chloride (NH4Cl) and place it in a100-mL beaker

2 Add 30 mL of water to the NH4Cl stirring with your stirring rod

3 Feel the bottom of the beaker and record your observations

4 Repeat the procedure with calcium chloride (CaCl2)

Analysis

Which dissolving process is exothermic Endothermic Suggest somepractical applications for dissolving processes that are exothermicand for those that are endothermic

Objectives bull Describe the characteristics

of solutions and identify thevarious types

bull Relate intermolecular forcesand the process of solva-tion

bull Define solubility and iden-tify factors affecting it

Vocabularysolubleinsolubleimmisciblemisciblesolvationheat of solutionsolubilitysaturated solutionunsaturated solutionsupersaturated solutionHenryrsquos law

Section 151 What are solutions

Have you ever thought of the importance of solutions Even if you havenrsquotthe fact is that solutions are all around you They are even inside you youand all other organisms are composed of cells containing solutions that sup-port life You inhale a solution when you breathe You are immersed in a solu-tion whether standing in a room or swimming in a pool Structures such asthe ones shown in the photo on the previous page would not be possible with-out steel yet another solution

Characteristics of SolutionsCell solutions ocean water and steel may appear quite dissimilar but theyshare certain characteristics In Chapter 3 you learned that solutions arehomogeneous mixtures containing two or more substances called the soluteand the solvent The solute is the substance that dissolves The solvent is thedissolving medium When you look at a solution it is not possible to distin-guish the solute from the solvent

A solution may exist as a gas liquid or solid depending on the state of itssolvent as shown in Figure 15-1 and Table 15-1 on the next page Air is agaseous solution and its solvent is nitrogen gas braces may be made of niti-nol a solid solution of titanium in nickel Most solutions however are liq-uids You learned in Chapter 10 that reactions can take place in aqueoussolutions that is solutions in which reactants and products are mixed inwater In fact water is the most common solvent among liquid solutions

Chem MC-453

null

1044361

454 Chapter 15 Solutions

Table 15-1 lists examples of different types of solutions Note that thesolutes in the solutions may be gases liquids or solids Which solutions con-tain gaseous solutes Which solutions are aqueous Note also that solutionssuch as ocean water can contain more than one solute

Some combinations of substances readily form solutions and others do notA substance that dissolves in a solvent is said to be soluble in that solventFor example sugar is soluble in water a fact you probably learned by dis-solving sugar in flavored water to make a sweetened beverage such as tealemonade or fruit punch A substance that does not dissolve in a solvent issaid to be insoluble in that solvent Sand is insoluble in water Have you evershaken a bottle of oil and vinegar when making salad dressing If so whathappens to the liquids shortly after you stop mixing them You are correctif you answered that they separate or cease to mix Oil is insoluble in vine-gar and thus oil and vinegar are said to be immiscible Two liquids that aresoluble in each other such as those that form the antifreeze listed inTable 15-1 are said to be miscible Are water and acetic acid miscible

Figure 15-1

The air you breathe is a gassolution primarily containingoxygen nitrogen and argon

The biological reactions nec-essary for life occur in aqueoussolutions within cells A solidsolution of titanium and nickel iscommonly used for braces inorthodontia

c

b

a

Types and Examples of Solutions

Type of solution Example Solvent Solute

Gas

Gas in gas Air Nitrogen (gas) Oxygen (gas)

Liquid

Gas in liquid Carbonated water Water (liquid) Carbon dioxide (gas)

Gas in liquid Ocean water Water (liquid) Oxygen gas (gas)

Liquid in liquid Antifreeze Water (liquid) Ethylene glycol(liquid)

Liquid in liquid Vinegar Water (liquid) Acetic acid (liquid)

Solid in liquid Ocean water Water (liquid) Sodium chloride(solid)

Solid

Liquid in solid Dental amalgam Silver (solid) Mercury (liquid)

Solid in solid Steel Iron (solid) Carbon (solid)

Table 15-1

a b c

Chem MC-454

null

81188324

151 What are solutions 455

Solvation in Aqueous SolutionsWhy are some substances soluble in one another whereas others are not Toform a solution solute particles must separate from one another and thesolute and solvent particles must mix Recall from Chapter 13 that attractiveforces exist among the particles of all substances Attractive forces existbetween the pure solute particles between the pure solvent particles andbetween the solute and solvent particles When a solid solute is placed in asolvent the solvent particles completely surround the surface of the solidsolute If the attractive forces between the solvent and solute particles aregreater than the attractive forces holding the solute particles together the sol-vent particles pull the solute particles apart and surround them These sur-rounded solute particles then move away from the solid solute out into thesolution The process of surrounding solute particles with solvent particles toform a solution is called solvation Solvation in water is called hydration

ldquoLike dissolves likerdquo is the general rule used to determine whether solva-tion will occur in a specific solvent To determine whether a solvent and soluteare alike you must examine the bonding and the polarity of the particles andthe intermolecular forces between particles

Aqueous solutions of ionic compounds Examine Figure 15-2 to reviewthe polar nature of water Recall that water molecules are dipoles with par-tially positive and partially negative ends Water molecules are in constantmotion as described by the kineticndashmolecular theory When a crystal of anionic compound such as sodium chloride (NaCl) is placed in a beaker ofwater the water molecules collide with the surface of the crystal The chargedends of the water molecules attract the positive sodium ions and negative chlo-ride ions This attraction between the dipoles and the ions is greater than theattraction among the ions in the crystal and so the ions break away from thesurface The water molecules surround the ions and the solvated ions moveinto solution as shown in Figure 15-3 This exposes more ions on the sur-face of the crystal Solvation continues until the entire crystal has dissolvedand all ions are distributed throughout the solvent

Gypsum is a compound composed of calcium ions and sulfate ions It ismixed with water to make plaster Plaster is a mixture not a solution Gypsumis insoluble in water because the attractive forces among the ions in calciumsulfate are so strong that they cannot be overcome by the attractive forcesexerted by the water molecules Solvation does not occur

Figure 15-2

The bent shape of a water mole-cule results in dipoles that donot cancel each other out Themolecule has a net polarity withthe oxygen end being partiallynegative and the hydrogen endsbeing partially positive Refer toTable C-1 in Appendix C for akey to atom color conventions

Figure 15-3

Solid sodium chloride dissolvesas its ions are surrounded by sol-vent water molecules Note howthe polar water molecules orientthemselves differently aroundthe positive and negative ions

Na ions

Water molecules

Hydrated ions

Cl ions

Solvation Process of NaCl

Chem MC-455

null

1764326

Aqueous solutions of molecular compounds Water also is a good sol-vent for many molecular compounds You know that table sugar dissolves inwater Table sugar is the molecular compound sucrose See Figure 15-4Note that its structure has a number of OndashH bonds Sucrose molecules arepolar When water is added to sucrose each OndashH bond becomes a site forhydrogen bonding with water As soon as the sugar crystals contact the waterwater molecules collide with the outer surface of the crystal The attractiveforces among sucrose molecules are overcome by the attractive forcesbetween polar water molecules and polar sucrose molecules Sugar moleculesleave the crystal and become solvated by water molecules

The oil in Figure 15-5 is a substance made up of primarily carbon andhydrogen It does not form a solution with water Why are oil and water immis-cible There is little attraction between the polar water molecules and the non-polar oil molecules However oil spills can be cleaned up with a nonpolarsolvent Nonpolar solutes are more readily dissolved in nonpolar solvents

Factors that affect rate of solvation As you have just learned solva-tion occurs only when and where the solute and solvent particles come in con-tact with each other There are three common ways to increase the collisionsbetween solute and solvent particles and thus increase the rate at which thesolute dissolves agitating the mixture increasing the surface area of thesolute and increasing the temperature of the solvent

Agitating the mixture by stirring and shaking moves dissolved solute par-ticles away from the contact surfaces more quickly and thereby allows newcollisions between solute and solvent particles to occur Without stirring orshaking solvated particles move away from the contact areas slowly Breakingthe solute into small pieces increases its surface area A greater surface areaallows more collisions to occur This is why a teaspoon of granulated sugardissolves more quickly than an equal amount of sugar in a cube Raising thetemperature of the solvent increases the kinetic energy of its particles result-ing in more frequent collisions and collisions with greater energy than thosethat occur at lower temperatures Can you think of an example of how anincrease in temperature affects the rate of dissolving

456 Chapter 15 Solutions

Figure 15-4

The solvation of a sugar cube(sucrose) can be seen in theabove photo Polar sucrose(C12H22O11) molecules containeight OndashH bonds Forcesbetween polar water moleculesand polar sucrose moleculesbreak these OndashH bonds and thesucrose dissolves

C

H O

H H

mdashmdash

mdash

mdash

C

C

H

H

O mdash C

OC

H mdash O

O mdash C C

O

O mdash H

H

H mdash O

H

H mdash O

H

H

HO mdash HH

H

H mdash O

H

H mdash OHH

H

mdash

O H

H OCH2 mdash O

mdashmdash

mdashmdash

O

H mdash O

CH

mdashC

CH2

mdash

CH2 O mdash H

H

H mdash O

mdash

H

mdash

H

mdash

Figure 15-5

Nonpolar oil molecules do notmix with polar water moleculesBecause oil is less dense thanwater it floats on the waterrsquossurface For this reason oil spillsat sea often wash ashore

Chem MC-456

null

1498896

Heat of solution During the process of solvation the solute must separateinto particles Solvent particles also must move apart in order to allow soluteparticles to come between them Energy is required to overcome the attrac-tive forces within the solute and within the solvent so both steps are endother-mic When solute and solvent particles mix the particles attract each otherand energy is released This step in the solvation process is exothermic Theoverall energy change that occurs during the solution formation process iscalled the heat of solution

As you observed in the DISCOVERY LAB at the beginning of this chap-ter some solutions release energy as they form whereas others absorb energyduring formation For example after ammonium nitrate dissolves in waterits container feels cool In contrast after calcium chloride dissolves in waterits container feels warm You will learn more about the heat of solution in thenext chapter

SolubilityIf you have ever added so much sugar to a sweetened beverage that sugar crys-tals accumulated on the containerrsquos bottom then you know that only a lim-ited amount of solute can dissolve in a solvent at a given set of conditionsIn fact every solute has a characteristic solubility Solubility refers to the max-imum amount of solute that will dissolve in a given amount of solvent at aspecified temperature and pressure As you can see from Table 15-2 solu-bility is usually expressed in grams of solute per 100 g of solvent

151 What are solutions 457

Solubilities of Some Solutes in Water at Various Temperatures

Substance Formula Solubility (g100 g H2O)

0degC 20degC 60degC 100degC

Aluminum sulfate Al2(SO4)3 312 364 592 890

Ammonium chloride NH4Cl 294 372 553 773

Barium hydroxide Ba(OH)2 167 389 2094 mdash

Barium nitrate Ba(NO3)2 495 902 204 344

Calcium hydroxide Ca(OH)2 0189 0173 0121 0076

Lead(II) chloride PbCl2 067 100 194 320

Lithium sulfate Li2SO4 361 348 326 mdash

Potassium chloride KCl 280 342 458 563

Potassium sulfate K2SO4 74 111 182 241

Sodium chloride NaCl 357 359 371 392

Silver nitrate AgNO3 122 216 440 733

Sucrose C12H22O11 1792 2039 2873 4872

Ammonia NH3 1130 680 200 mdash

Carbon dioxide CO2 1713 0878 0359 mdash

Oxygen O2 0048 0031 0019 mdash

L1 L H2O of gas at standard pressure (101 kPa) mdash No value available

Table 15-2

Earth ScienceCONNECTION

Water is the most abundantsolvent on Earth The

amount of oxygen contained inwater is called dissolved oxygen(DO) Dissolved oxygen is an indi-cator of water quality becauseoxygen is necessary for fish andother aquatic life Oxygen entersa body of water by photosynthe-sis of aquatic plants and by trans-fer of oxygen across the air-waterboundary

Thermal pollution is a reduc-tion in water quality due to anincrease in water temperatureThe discharge of heated waterinto a lake river or other body ofwater by factories or powerplants causes the solubility ofoxygen in water to decrease Inaddition to not having enoughoxygen to support life as shownbelow the lower oxygen levelmagnifies the effects of toxic andorganic pollutants

To reduce thermal pollutionmeasures must be taken beforethe heated water is released intothe environment The water canbe discharged into holding lakesor canals and allowed to cool Inaddition many facilities use cool-ing towers to dissipate heat intothe air

Chem MC-457

null

96913216

Just as solvation can be understood at the par-ticle level so can solubility When a solute isadded to a solvent solvent particles collide withthe solutersquos surface particles solute particles beginto mix randomly among the solvent particles Atfirst the solute particles are carried away from thecrystal However as the number of solvated par-ticles increases the same random mixing results inincreasingly frequent collisions between solvatedsolute particles and the remaining crystal Somecolliding solute particles rejoin the crystal or crys-tallize as you can see in Figure 15-6 As solvationcontinues the crystallization rate increases whilethe solvation rate remains constant As long as thesolvation rate is greater than the crystallizationrate the net effect is continuing solvation

Depending on the amount of solute present the rates of solvation andcrystallization may eventually equalize no more solute appears to dissolveand a state of dynamic equilibrium exists between crystallization and solva-tion (as long as the temperature remains constant) Although solute particlescontinue to dissolve and crystallize in solutions that reach equilibrium theoverall amount of dissolved solute in the solution remains constant Such asolution is said to be a saturated solution it contains the maximum amountof dissolved solute for a given amount of solvent at a specific temperatureand pressure An unsaturated solution is one that contains less dissolvedsolute for a given temperature and pressure than a saturated solution In otherwords more solute can be dissolved in an unsaturated solution

Factors That Affect SolubilityPressure affects the solubility of gaseous solutes and gaseous solutions Thesolubility of a solute also depends on the nature of the solute and solventTemperature affects the solubility of all substances To learn how a blood dis-order called sickle-cell disease can affect oxygenrsquos solubility in blood readthe Chemistry and Society feature at the end of this chapter

Temperature and solubility Many substances aremore soluble at high temperatures than at low tempera-tures as you can see by the data graphed in Figure 15-7For example calcium chloride (CaCl2) has a solubility ofabout 64 g CaCl2 per 100 g H2O at 10degC Increasing thetemperature to approximately 27degC increases the solu-bility by 50 to 100 g CaCl2 per 100 g H2O This factalso is illustrated by the data in Table 15-2 on the previ-ous page From the table you can see that at 20degC 2039 gof sucrose (C12H22O11) dissolves in 100 g of water At100degC 4872 g of sucrose dissolves in 100 g of waternearly a 140 increase in solubility over an 80degC tem-perature range According to Figure 15-7 and Table 15-2what substances decrease in solubility as temperatureincreases

458 Chapter 15 Solutions

0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g o

f so

lute

100

g H

2O) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

CrystallizationSolvation

Figure 15-6

A dynamic equilibrium exists in asaturated solution That is therate at which solute particles inthe crystal are solvated is equalto the rate at which solvatedsolute particles rejoin the crystal

Figure 15-7

This graph shows the solubilityof several substances as a func-tion of temperature What is thesolubility of KClO3 at 60degC

Chem MC-458

null

20219186

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

454 Chapter 15 Solutions

Table 15-1 lists examples of different types of solutions Note that thesolutes in the solutions may be gases liquids or solids Which solutions con-tain gaseous solutes Which solutions are aqueous Note also that solutionssuch as ocean water can contain more than one solute

Some combinations of substances readily form solutions and others do notA substance that dissolves in a solvent is said to be soluble in that solventFor example sugar is soluble in water a fact you probably learned by dis-solving sugar in flavored water to make a sweetened beverage such as tealemonade or fruit punch A substance that does not dissolve in a solvent issaid to be insoluble in that solvent Sand is insoluble in water Have you evershaken a bottle of oil and vinegar when making salad dressing If so whathappens to the liquids shortly after you stop mixing them You are correctif you answered that they separate or cease to mix Oil is insoluble in vine-gar and thus oil and vinegar are said to be immiscible Two liquids that aresoluble in each other such as those that form the antifreeze listed inTable 15-1 are said to be miscible Are water and acetic acid miscible

Figure 15-1

The air you breathe is a gassolution primarily containingoxygen nitrogen and argon

The biological reactions nec-essary for life occur in aqueoussolutions within cells A solidsolution of titanium and nickel iscommonly used for braces inorthodontia

c

b

a

Types and Examples of Solutions

Type of solution Example Solvent Solute

Gas

Gas in gas Air Nitrogen (gas) Oxygen (gas)

Liquid

Gas in liquid Carbonated water Water (liquid) Carbon dioxide (gas)

Gas in liquid Ocean water Water (liquid) Oxygen gas (gas)

Liquid in liquid Antifreeze Water (liquid) Ethylene glycol(liquid)

Liquid in liquid Vinegar Water (liquid) Acetic acid (liquid)

Solid in liquid Ocean water Water (liquid) Sodium chloride(solid)

Solid

Liquid in solid Dental amalgam Silver (solid) Mercury (liquid)

Solid in solid Steel Iron (solid) Carbon (solid)

Table 15-1

a b c

Chem MC-454

null

81188324

151 What are solutions 455

Solvation in Aqueous SolutionsWhy are some substances soluble in one another whereas others are not Toform a solution solute particles must separate from one another and thesolute and solvent particles must mix Recall from Chapter 13 that attractiveforces exist among the particles of all substances Attractive forces existbetween the pure solute particles between the pure solvent particles andbetween the solute and solvent particles When a solid solute is placed in asolvent the solvent particles completely surround the surface of the solidsolute If the attractive forces between the solvent and solute particles aregreater than the attractive forces holding the solute particles together the sol-vent particles pull the solute particles apart and surround them These sur-rounded solute particles then move away from the solid solute out into thesolution The process of surrounding solute particles with solvent particles toform a solution is called solvation Solvation in water is called hydration

ldquoLike dissolves likerdquo is the general rule used to determine whether solva-tion will occur in a specific solvent To determine whether a solvent and soluteare alike you must examine the bonding and the polarity of the particles andthe intermolecular forces between particles

Aqueous solutions of ionic compounds Examine Figure 15-2 to reviewthe polar nature of water Recall that water molecules are dipoles with par-tially positive and partially negative ends Water molecules are in constantmotion as described by the kineticndashmolecular theory When a crystal of anionic compound such as sodium chloride (NaCl) is placed in a beaker ofwater the water molecules collide with the surface of the crystal The chargedends of the water molecules attract the positive sodium ions and negative chlo-ride ions This attraction between the dipoles and the ions is greater than theattraction among the ions in the crystal and so the ions break away from thesurface The water molecules surround the ions and the solvated ions moveinto solution as shown in Figure 15-3 This exposes more ions on the sur-face of the crystal Solvation continues until the entire crystal has dissolvedand all ions are distributed throughout the solvent

Gypsum is a compound composed of calcium ions and sulfate ions It ismixed with water to make plaster Plaster is a mixture not a solution Gypsumis insoluble in water because the attractive forces among the ions in calciumsulfate are so strong that they cannot be overcome by the attractive forcesexerted by the water molecules Solvation does not occur

Figure 15-2

The bent shape of a water mole-cule results in dipoles that donot cancel each other out Themolecule has a net polarity withthe oxygen end being partiallynegative and the hydrogen endsbeing partially positive Refer toTable C-1 in Appendix C for akey to atom color conventions

Figure 15-3

Solid sodium chloride dissolvesas its ions are surrounded by sol-vent water molecules Note howthe polar water molecules orientthemselves differently aroundthe positive and negative ions

Na ions

Water molecules

Hydrated ions

Cl ions

Solvation Process of NaCl

Chem MC-455

null

1764326

Aqueous solutions of molecular compounds Water also is a good sol-vent for many molecular compounds You know that table sugar dissolves inwater Table sugar is the molecular compound sucrose See Figure 15-4Note that its structure has a number of OndashH bonds Sucrose molecules arepolar When water is added to sucrose each OndashH bond becomes a site forhydrogen bonding with water As soon as the sugar crystals contact the waterwater molecules collide with the outer surface of the crystal The attractiveforces among sucrose molecules are overcome by the attractive forcesbetween polar water molecules and polar sucrose molecules Sugar moleculesleave the crystal and become solvated by water molecules

The oil in Figure 15-5 is a substance made up of primarily carbon andhydrogen It does not form a solution with water Why are oil and water immis-cible There is little attraction between the polar water molecules and the non-polar oil molecules However oil spills can be cleaned up with a nonpolarsolvent Nonpolar solutes are more readily dissolved in nonpolar solvents

Factors that affect rate of solvation As you have just learned solva-tion occurs only when and where the solute and solvent particles come in con-tact with each other There are three common ways to increase the collisionsbetween solute and solvent particles and thus increase the rate at which thesolute dissolves agitating the mixture increasing the surface area of thesolute and increasing the temperature of the solvent

Agitating the mixture by stirring and shaking moves dissolved solute par-ticles away from the contact surfaces more quickly and thereby allows newcollisions between solute and solvent particles to occur Without stirring orshaking solvated particles move away from the contact areas slowly Breakingthe solute into small pieces increases its surface area A greater surface areaallows more collisions to occur This is why a teaspoon of granulated sugardissolves more quickly than an equal amount of sugar in a cube Raising thetemperature of the solvent increases the kinetic energy of its particles result-ing in more frequent collisions and collisions with greater energy than thosethat occur at lower temperatures Can you think of an example of how anincrease in temperature affects the rate of dissolving

456 Chapter 15 Solutions

Figure 15-4

The solvation of a sugar cube(sucrose) can be seen in theabove photo Polar sucrose(C12H22O11) molecules containeight OndashH bonds Forcesbetween polar water moleculesand polar sucrose moleculesbreak these OndashH bonds and thesucrose dissolves

C

H O

H H

mdashmdash

mdash

mdash

C

C

H

H

O mdash C

OC

H mdash O

O mdash C C

O

O mdash H

H

H mdash O

H

H mdash O

H

H

HO mdash HH

H

H mdash O

H

H mdash OHH

H

mdash

O H

H OCH2 mdash O

mdashmdash

mdashmdash

O

H mdash O

CH

mdashC

CH2

mdash

CH2 O mdash H

H

H mdash O

mdash

H

mdash

H

mdash

Figure 15-5

Nonpolar oil molecules do notmix with polar water moleculesBecause oil is less dense thanwater it floats on the waterrsquossurface For this reason oil spillsat sea often wash ashore

Chem MC-456

null

1498896

Heat of solution During the process of solvation the solute must separateinto particles Solvent particles also must move apart in order to allow soluteparticles to come between them Energy is required to overcome the attrac-tive forces within the solute and within the solvent so both steps are endother-mic When solute and solvent particles mix the particles attract each otherand energy is released This step in the solvation process is exothermic Theoverall energy change that occurs during the solution formation process iscalled the heat of solution

As you observed in the DISCOVERY LAB at the beginning of this chap-ter some solutions release energy as they form whereas others absorb energyduring formation For example after ammonium nitrate dissolves in waterits container feels cool In contrast after calcium chloride dissolves in waterits container feels warm You will learn more about the heat of solution in thenext chapter

SolubilityIf you have ever added so much sugar to a sweetened beverage that sugar crys-tals accumulated on the containerrsquos bottom then you know that only a lim-ited amount of solute can dissolve in a solvent at a given set of conditionsIn fact every solute has a characteristic solubility Solubility refers to the max-imum amount of solute that will dissolve in a given amount of solvent at aspecified temperature and pressure As you can see from Table 15-2 solu-bility is usually expressed in grams of solute per 100 g of solvent

151 What are solutions 457

Solubilities of Some Solutes in Water at Various Temperatures

Substance Formula Solubility (g100 g H2O)

0degC 20degC 60degC 100degC

Aluminum sulfate Al2(SO4)3 312 364 592 890

Ammonium chloride NH4Cl 294 372 553 773

Barium hydroxide Ba(OH)2 167 389 2094 mdash

Barium nitrate Ba(NO3)2 495 902 204 344

Calcium hydroxide Ca(OH)2 0189 0173 0121 0076

Lead(II) chloride PbCl2 067 100 194 320

Lithium sulfate Li2SO4 361 348 326 mdash

Potassium chloride KCl 280 342 458 563

Potassium sulfate K2SO4 74 111 182 241

Sodium chloride NaCl 357 359 371 392

Silver nitrate AgNO3 122 216 440 733

Sucrose C12H22O11 1792 2039 2873 4872

Ammonia NH3 1130 680 200 mdash

Carbon dioxide CO2 1713 0878 0359 mdash

Oxygen O2 0048 0031 0019 mdash

L1 L H2O of gas at standard pressure (101 kPa) mdash No value available

Table 15-2

Earth ScienceCONNECTION

Water is the most abundantsolvent on Earth The

amount of oxygen contained inwater is called dissolved oxygen(DO) Dissolved oxygen is an indi-cator of water quality becauseoxygen is necessary for fish andother aquatic life Oxygen entersa body of water by photosynthe-sis of aquatic plants and by trans-fer of oxygen across the air-waterboundary

Thermal pollution is a reduc-tion in water quality due to anincrease in water temperatureThe discharge of heated waterinto a lake river or other body ofwater by factories or powerplants causes the solubility ofoxygen in water to decrease Inaddition to not having enoughoxygen to support life as shownbelow the lower oxygen levelmagnifies the effects of toxic andorganic pollutants

To reduce thermal pollutionmeasures must be taken beforethe heated water is released intothe environment The water canbe discharged into holding lakesor canals and allowed to cool Inaddition many facilities use cool-ing towers to dissipate heat intothe air

Chem MC-457

null

96913216

Just as solvation can be understood at the par-ticle level so can solubility When a solute isadded to a solvent solvent particles collide withthe solutersquos surface particles solute particles beginto mix randomly among the solvent particles Atfirst the solute particles are carried away from thecrystal However as the number of solvated par-ticles increases the same random mixing results inincreasingly frequent collisions between solvatedsolute particles and the remaining crystal Somecolliding solute particles rejoin the crystal or crys-tallize as you can see in Figure 15-6 As solvationcontinues the crystallization rate increases whilethe solvation rate remains constant As long as thesolvation rate is greater than the crystallizationrate the net effect is continuing solvation

Depending on the amount of solute present the rates of solvation andcrystallization may eventually equalize no more solute appears to dissolveand a state of dynamic equilibrium exists between crystallization and solva-tion (as long as the temperature remains constant) Although solute particlescontinue to dissolve and crystallize in solutions that reach equilibrium theoverall amount of dissolved solute in the solution remains constant Such asolution is said to be a saturated solution it contains the maximum amountof dissolved solute for a given amount of solvent at a specific temperatureand pressure An unsaturated solution is one that contains less dissolvedsolute for a given temperature and pressure than a saturated solution In otherwords more solute can be dissolved in an unsaturated solution

Factors That Affect SolubilityPressure affects the solubility of gaseous solutes and gaseous solutions Thesolubility of a solute also depends on the nature of the solute and solventTemperature affects the solubility of all substances To learn how a blood dis-order called sickle-cell disease can affect oxygenrsquos solubility in blood readthe Chemistry and Society feature at the end of this chapter

Temperature and solubility Many substances aremore soluble at high temperatures than at low tempera-tures as you can see by the data graphed in Figure 15-7For example calcium chloride (CaCl2) has a solubility ofabout 64 g CaCl2 per 100 g H2O at 10degC Increasing thetemperature to approximately 27degC increases the solu-bility by 50 to 100 g CaCl2 per 100 g H2O This factalso is illustrated by the data in Table 15-2 on the previ-ous page From the table you can see that at 20degC 2039 gof sucrose (C12H22O11) dissolves in 100 g of water At100degC 4872 g of sucrose dissolves in 100 g of waternearly a 140 increase in solubility over an 80degC tem-perature range According to Figure 15-7 and Table 15-2what substances decrease in solubility as temperatureincreases

458 Chapter 15 Solutions

0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g o

f so

lute

100

g H

2O) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

CrystallizationSolvation

Figure 15-6

A dynamic equilibrium exists in asaturated solution That is therate at which solute particles inthe crystal are solvated is equalto the rate at which solvatedsolute particles rejoin the crystal

Figure 15-7

This graph shows the solubilityof several substances as a func-tion of temperature What is thesolubility of KClO3 at 60degC

Chem MC-458

null

20219186

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

151 What are solutions 455

Solvation in Aqueous SolutionsWhy are some substances soluble in one another whereas others are not Toform a solution solute particles must separate from one another and thesolute and solvent particles must mix Recall from Chapter 13 that attractiveforces exist among the particles of all substances Attractive forces existbetween the pure solute particles between the pure solvent particles andbetween the solute and solvent particles When a solid solute is placed in asolvent the solvent particles completely surround the surface of the solidsolute If the attractive forces between the solvent and solute particles aregreater than the attractive forces holding the solute particles together the sol-vent particles pull the solute particles apart and surround them These sur-rounded solute particles then move away from the solid solute out into thesolution The process of surrounding solute particles with solvent particles toform a solution is called solvation Solvation in water is called hydration

ldquoLike dissolves likerdquo is the general rule used to determine whether solva-tion will occur in a specific solvent To determine whether a solvent and soluteare alike you must examine the bonding and the polarity of the particles andthe intermolecular forces between particles

Aqueous solutions of ionic compounds Examine Figure 15-2 to reviewthe polar nature of water Recall that water molecules are dipoles with par-tially positive and partially negative ends Water molecules are in constantmotion as described by the kineticndashmolecular theory When a crystal of anionic compound such as sodium chloride (NaCl) is placed in a beaker ofwater the water molecules collide with the surface of the crystal The chargedends of the water molecules attract the positive sodium ions and negative chlo-ride ions This attraction between the dipoles and the ions is greater than theattraction among the ions in the crystal and so the ions break away from thesurface The water molecules surround the ions and the solvated ions moveinto solution as shown in Figure 15-3 This exposes more ions on the sur-face of the crystal Solvation continues until the entire crystal has dissolvedand all ions are distributed throughout the solvent

Gypsum is a compound composed of calcium ions and sulfate ions It ismixed with water to make plaster Plaster is a mixture not a solution Gypsumis insoluble in water because the attractive forces among the ions in calciumsulfate are so strong that they cannot be overcome by the attractive forcesexerted by the water molecules Solvation does not occur

Figure 15-2

The bent shape of a water mole-cule results in dipoles that donot cancel each other out Themolecule has a net polarity withthe oxygen end being partiallynegative and the hydrogen endsbeing partially positive Refer toTable C-1 in Appendix C for akey to atom color conventions

Figure 15-3

Solid sodium chloride dissolvesas its ions are surrounded by sol-vent water molecules Note howthe polar water molecules orientthemselves differently aroundthe positive and negative ions

Na ions

Water molecules

Hydrated ions

Cl ions

Solvation Process of NaCl

Chem MC-455

null

1764326

Aqueous solutions of molecular compounds Water also is a good sol-vent for many molecular compounds You know that table sugar dissolves inwater Table sugar is the molecular compound sucrose See Figure 15-4Note that its structure has a number of OndashH bonds Sucrose molecules arepolar When water is added to sucrose each OndashH bond becomes a site forhydrogen bonding with water As soon as the sugar crystals contact the waterwater molecules collide with the outer surface of the crystal The attractiveforces among sucrose molecules are overcome by the attractive forcesbetween polar water molecules and polar sucrose molecules Sugar moleculesleave the crystal and become solvated by water molecules

The oil in Figure 15-5 is a substance made up of primarily carbon andhydrogen It does not form a solution with water Why are oil and water immis-cible There is little attraction between the polar water molecules and the non-polar oil molecules However oil spills can be cleaned up with a nonpolarsolvent Nonpolar solutes are more readily dissolved in nonpolar solvents

Factors that affect rate of solvation As you have just learned solva-tion occurs only when and where the solute and solvent particles come in con-tact with each other There are three common ways to increase the collisionsbetween solute and solvent particles and thus increase the rate at which thesolute dissolves agitating the mixture increasing the surface area of thesolute and increasing the temperature of the solvent

Agitating the mixture by stirring and shaking moves dissolved solute par-ticles away from the contact surfaces more quickly and thereby allows newcollisions between solute and solvent particles to occur Without stirring orshaking solvated particles move away from the contact areas slowly Breakingthe solute into small pieces increases its surface area A greater surface areaallows more collisions to occur This is why a teaspoon of granulated sugardissolves more quickly than an equal amount of sugar in a cube Raising thetemperature of the solvent increases the kinetic energy of its particles result-ing in more frequent collisions and collisions with greater energy than thosethat occur at lower temperatures Can you think of an example of how anincrease in temperature affects the rate of dissolving

456 Chapter 15 Solutions

Figure 15-4

The solvation of a sugar cube(sucrose) can be seen in theabove photo Polar sucrose(C12H22O11) molecules containeight OndashH bonds Forcesbetween polar water moleculesand polar sucrose moleculesbreak these OndashH bonds and thesucrose dissolves

C

H O

H H

mdashmdash

mdash

mdash

C

C

H

H

O mdash C

OC

H mdash O

O mdash C C

O

O mdash H

H

H mdash O

H

H mdash O

H

H

HO mdash HH

H

H mdash O

H

H mdash OHH

H

mdash

O H

H OCH2 mdash O

mdashmdash

mdashmdash

O

H mdash O

CH

mdashC

CH2

mdash

CH2 O mdash H

H

H mdash O

mdash

H

mdash

H

mdash

Figure 15-5

Nonpolar oil molecules do notmix with polar water moleculesBecause oil is less dense thanwater it floats on the waterrsquossurface For this reason oil spillsat sea often wash ashore

Chem MC-456

null

1498896

Heat of solution During the process of solvation the solute must separateinto particles Solvent particles also must move apart in order to allow soluteparticles to come between them Energy is required to overcome the attrac-tive forces within the solute and within the solvent so both steps are endother-mic When solute and solvent particles mix the particles attract each otherand energy is released This step in the solvation process is exothermic Theoverall energy change that occurs during the solution formation process iscalled the heat of solution

As you observed in the DISCOVERY LAB at the beginning of this chap-ter some solutions release energy as they form whereas others absorb energyduring formation For example after ammonium nitrate dissolves in waterits container feels cool In contrast after calcium chloride dissolves in waterits container feels warm You will learn more about the heat of solution in thenext chapter

SolubilityIf you have ever added so much sugar to a sweetened beverage that sugar crys-tals accumulated on the containerrsquos bottom then you know that only a lim-ited amount of solute can dissolve in a solvent at a given set of conditionsIn fact every solute has a characteristic solubility Solubility refers to the max-imum amount of solute that will dissolve in a given amount of solvent at aspecified temperature and pressure As you can see from Table 15-2 solu-bility is usually expressed in grams of solute per 100 g of solvent

151 What are solutions 457

Solubilities of Some Solutes in Water at Various Temperatures

Substance Formula Solubility (g100 g H2O)

0degC 20degC 60degC 100degC

Aluminum sulfate Al2(SO4)3 312 364 592 890

Ammonium chloride NH4Cl 294 372 553 773

Barium hydroxide Ba(OH)2 167 389 2094 mdash

Barium nitrate Ba(NO3)2 495 902 204 344

Calcium hydroxide Ca(OH)2 0189 0173 0121 0076

Lead(II) chloride PbCl2 067 100 194 320

Lithium sulfate Li2SO4 361 348 326 mdash

Potassium chloride KCl 280 342 458 563

Potassium sulfate K2SO4 74 111 182 241

Sodium chloride NaCl 357 359 371 392

Silver nitrate AgNO3 122 216 440 733

Sucrose C12H22O11 1792 2039 2873 4872

Ammonia NH3 1130 680 200 mdash

Carbon dioxide CO2 1713 0878 0359 mdash

Oxygen O2 0048 0031 0019 mdash

L1 L H2O of gas at standard pressure (101 kPa) mdash No value available

Table 15-2

Earth ScienceCONNECTION

Water is the most abundantsolvent on Earth The

amount of oxygen contained inwater is called dissolved oxygen(DO) Dissolved oxygen is an indi-cator of water quality becauseoxygen is necessary for fish andother aquatic life Oxygen entersa body of water by photosynthe-sis of aquatic plants and by trans-fer of oxygen across the air-waterboundary

Thermal pollution is a reduc-tion in water quality due to anincrease in water temperatureThe discharge of heated waterinto a lake river or other body ofwater by factories or powerplants causes the solubility ofoxygen in water to decrease Inaddition to not having enoughoxygen to support life as shownbelow the lower oxygen levelmagnifies the effects of toxic andorganic pollutants

To reduce thermal pollutionmeasures must be taken beforethe heated water is released intothe environment The water canbe discharged into holding lakesor canals and allowed to cool Inaddition many facilities use cool-ing towers to dissipate heat intothe air

Chem MC-457

null

96913216

Just as solvation can be understood at the par-ticle level so can solubility When a solute isadded to a solvent solvent particles collide withthe solutersquos surface particles solute particles beginto mix randomly among the solvent particles Atfirst the solute particles are carried away from thecrystal However as the number of solvated par-ticles increases the same random mixing results inincreasingly frequent collisions between solvatedsolute particles and the remaining crystal Somecolliding solute particles rejoin the crystal or crys-tallize as you can see in Figure 15-6 As solvationcontinues the crystallization rate increases whilethe solvation rate remains constant As long as thesolvation rate is greater than the crystallizationrate the net effect is continuing solvation

Depending on the amount of solute present the rates of solvation andcrystallization may eventually equalize no more solute appears to dissolveand a state of dynamic equilibrium exists between crystallization and solva-tion (as long as the temperature remains constant) Although solute particlescontinue to dissolve and crystallize in solutions that reach equilibrium theoverall amount of dissolved solute in the solution remains constant Such asolution is said to be a saturated solution it contains the maximum amountof dissolved solute for a given amount of solvent at a specific temperatureand pressure An unsaturated solution is one that contains less dissolvedsolute for a given temperature and pressure than a saturated solution In otherwords more solute can be dissolved in an unsaturated solution

Factors That Affect SolubilityPressure affects the solubility of gaseous solutes and gaseous solutions Thesolubility of a solute also depends on the nature of the solute and solventTemperature affects the solubility of all substances To learn how a blood dis-order called sickle-cell disease can affect oxygenrsquos solubility in blood readthe Chemistry and Society feature at the end of this chapter

Temperature and solubility Many substances aremore soluble at high temperatures than at low tempera-tures as you can see by the data graphed in Figure 15-7For example calcium chloride (CaCl2) has a solubility ofabout 64 g CaCl2 per 100 g H2O at 10degC Increasing thetemperature to approximately 27degC increases the solu-bility by 50 to 100 g CaCl2 per 100 g H2O This factalso is illustrated by the data in Table 15-2 on the previ-ous page From the table you can see that at 20degC 2039 gof sucrose (C12H22O11) dissolves in 100 g of water At100degC 4872 g of sucrose dissolves in 100 g of waternearly a 140 increase in solubility over an 80degC tem-perature range According to Figure 15-7 and Table 15-2what substances decrease in solubility as temperatureincreases

458 Chapter 15 Solutions

0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g o

f so

lute

100

g H

2O) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

CrystallizationSolvation

Figure 15-6

A dynamic equilibrium exists in asaturated solution That is therate at which solute particles inthe crystal are solvated is equalto the rate at which solvatedsolute particles rejoin the crystal

Figure 15-7

This graph shows the solubilityof several substances as a func-tion of temperature What is thesolubility of KClO3 at 60degC

Chem MC-458

null

20219186

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

Aqueous solutions of molecular compounds Water also is a good sol-vent for many molecular compounds You know that table sugar dissolves inwater Table sugar is the molecular compound sucrose See Figure 15-4Note that its structure has a number of OndashH bonds Sucrose molecules arepolar When water is added to sucrose each OndashH bond becomes a site forhydrogen bonding with water As soon as the sugar crystals contact the waterwater molecules collide with the outer surface of the crystal The attractiveforces among sucrose molecules are overcome by the attractive forcesbetween polar water molecules and polar sucrose molecules Sugar moleculesleave the crystal and become solvated by water molecules

The oil in Figure 15-5 is a substance made up of primarily carbon andhydrogen It does not form a solution with water Why are oil and water immis-cible There is little attraction between the polar water molecules and the non-polar oil molecules However oil spills can be cleaned up with a nonpolarsolvent Nonpolar solutes are more readily dissolved in nonpolar solvents

Factors that affect rate of solvation As you have just learned solva-tion occurs only when and where the solute and solvent particles come in con-tact with each other There are three common ways to increase the collisionsbetween solute and solvent particles and thus increase the rate at which thesolute dissolves agitating the mixture increasing the surface area of thesolute and increasing the temperature of the solvent

Agitating the mixture by stirring and shaking moves dissolved solute par-ticles away from the contact surfaces more quickly and thereby allows newcollisions between solute and solvent particles to occur Without stirring orshaking solvated particles move away from the contact areas slowly Breakingthe solute into small pieces increases its surface area A greater surface areaallows more collisions to occur This is why a teaspoon of granulated sugardissolves more quickly than an equal amount of sugar in a cube Raising thetemperature of the solvent increases the kinetic energy of its particles result-ing in more frequent collisions and collisions with greater energy than thosethat occur at lower temperatures Can you think of an example of how anincrease in temperature affects the rate of dissolving

456 Chapter 15 Solutions

Figure 15-4

The solvation of a sugar cube(sucrose) can be seen in theabove photo Polar sucrose(C12H22O11) molecules containeight OndashH bonds Forcesbetween polar water moleculesand polar sucrose moleculesbreak these OndashH bonds and thesucrose dissolves

C

H O

H H

mdashmdash

mdash

mdash

C

C

H

H

O mdash C

OC

H mdash O

O mdash C C

O

O mdash H

H

H mdash O

H

H mdash O

H

H

HO mdash HH

H

H mdash O

H

H mdash OHH

H

mdash

O H

H OCH2 mdash O

mdashmdash

mdashmdash

O

H mdash O

CH

mdashC

CH2

mdash

CH2 O mdash H

H

H mdash O

mdash

H

mdash

H

mdash

Figure 15-5

Nonpolar oil molecules do notmix with polar water moleculesBecause oil is less dense thanwater it floats on the waterrsquossurface For this reason oil spillsat sea often wash ashore

Chem MC-456

null

1498896

Heat of solution During the process of solvation the solute must separateinto particles Solvent particles also must move apart in order to allow soluteparticles to come between them Energy is required to overcome the attrac-tive forces within the solute and within the solvent so both steps are endother-mic When solute and solvent particles mix the particles attract each otherand energy is released This step in the solvation process is exothermic Theoverall energy change that occurs during the solution formation process iscalled the heat of solution

As you observed in the DISCOVERY LAB at the beginning of this chap-ter some solutions release energy as they form whereas others absorb energyduring formation For example after ammonium nitrate dissolves in waterits container feels cool In contrast after calcium chloride dissolves in waterits container feels warm You will learn more about the heat of solution in thenext chapter

SolubilityIf you have ever added so much sugar to a sweetened beverage that sugar crys-tals accumulated on the containerrsquos bottom then you know that only a lim-ited amount of solute can dissolve in a solvent at a given set of conditionsIn fact every solute has a characteristic solubility Solubility refers to the max-imum amount of solute that will dissolve in a given amount of solvent at aspecified temperature and pressure As you can see from Table 15-2 solu-bility is usually expressed in grams of solute per 100 g of solvent

151 What are solutions 457

Solubilities of Some Solutes in Water at Various Temperatures

Substance Formula Solubility (g100 g H2O)

0degC 20degC 60degC 100degC

Aluminum sulfate Al2(SO4)3 312 364 592 890

Ammonium chloride NH4Cl 294 372 553 773

Barium hydroxide Ba(OH)2 167 389 2094 mdash

Barium nitrate Ba(NO3)2 495 902 204 344

Calcium hydroxide Ca(OH)2 0189 0173 0121 0076

Lead(II) chloride PbCl2 067 100 194 320

Lithium sulfate Li2SO4 361 348 326 mdash

Potassium chloride KCl 280 342 458 563

Potassium sulfate K2SO4 74 111 182 241

Sodium chloride NaCl 357 359 371 392

Silver nitrate AgNO3 122 216 440 733

Sucrose C12H22O11 1792 2039 2873 4872

Ammonia NH3 1130 680 200 mdash

Carbon dioxide CO2 1713 0878 0359 mdash

Oxygen O2 0048 0031 0019 mdash

L1 L H2O of gas at standard pressure (101 kPa) mdash No value available

Table 15-2

Earth ScienceCONNECTION

Water is the most abundantsolvent on Earth The

amount of oxygen contained inwater is called dissolved oxygen(DO) Dissolved oxygen is an indi-cator of water quality becauseoxygen is necessary for fish andother aquatic life Oxygen entersa body of water by photosynthe-sis of aquatic plants and by trans-fer of oxygen across the air-waterboundary

Thermal pollution is a reduc-tion in water quality due to anincrease in water temperatureThe discharge of heated waterinto a lake river or other body ofwater by factories or powerplants causes the solubility ofoxygen in water to decrease Inaddition to not having enoughoxygen to support life as shownbelow the lower oxygen levelmagnifies the effects of toxic andorganic pollutants

To reduce thermal pollutionmeasures must be taken beforethe heated water is released intothe environment The water canbe discharged into holding lakesor canals and allowed to cool Inaddition many facilities use cool-ing towers to dissipate heat intothe air

Chem MC-457

null

96913216

Just as solvation can be understood at the par-ticle level so can solubility When a solute isadded to a solvent solvent particles collide withthe solutersquos surface particles solute particles beginto mix randomly among the solvent particles Atfirst the solute particles are carried away from thecrystal However as the number of solvated par-ticles increases the same random mixing results inincreasingly frequent collisions between solvatedsolute particles and the remaining crystal Somecolliding solute particles rejoin the crystal or crys-tallize as you can see in Figure 15-6 As solvationcontinues the crystallization rate increases whilethe solvation rate remains constant As long as thesolvation rate is greater than the crystallizationrate the net effect is continuing solvation

Depending on the amount of solute present the rates of solvation andcrystallization may eventually equalize no more solute appears to dissolveand a state of dynamic equilibrium exists between crystallization and solva-tion (as long as the temperature remains constant) Although solute particlescontinue to dissolve and crystallize in solutions that reach equilibrium theoverall amount of dissolved solute in the solution remains constant Such asolution is said to be a saturated solution it contains the maximum amountof dissolved solute for a given amount of solvent at a specific temperatureand pressure An unsaturated solution is one that contains less dissolvedsolute for a given temperature and pressure than a saturated solution In otherwords more solute can be dissolved in an unsaturated solution

Factors That Affect SolubilityPressure affects the solubility of gaseous solutes and gaseous solutions Thesolubility of a solute also depends on the nature of the solute and solventTemperature affects the solubility of all substances To learn how a blood dis-order called sickle-cell disease can affect oxygenrsquos solubility in blood readthe Chemistry and Society feature at the end of this chapter

Temperature and solubility Many substances aremore soluble at high temperatures than at low tempera-tures as you can see by the data graphed in Figure 15-7For example calcium chloride (CaCl2) has a solubility ofabout 64 g CaCl2 per 100 g H2O at 10degC Increasing thetemperature to approximately 27degC increases the solu-bility by 50 to 100 g CaCl2 per 100 g H2O This factalso is illustrated by the data in Table 15-2 on the previ-ous page From the table you can see that at 20degC 2039 gof sucrose (C12H22O11) dissolves in 100 g of water At100degC 4872 g of sucrose dissolves in 100 g of waternearly a 140 increase in solubility over an 80degC tem-perature range According to Figure 15-7 and Table 15-2what substances decrease in solubility as temperatureincreases

458 Chapter 15 Solutions

0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g o

f so

lute

100

g H

2O) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

CrystallizationSolvation

Figure 15-6

A dynamic equilibrium exists in asaturated solution That is therate at which solute particles inthe crystal are solvated is equalto the rate at which solvatedsolute particles rejoin the crystal

Figure 15-7

This graph shows the solubilityof several substances as a func-tion of temperature What is thesolubility of KClO3 at 60degC

Chem MC-458

null

20219186

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

Heat of solution During the process of solvation the solute must separateinto particles Solvent particles also must move apart in order to allow soluteparticles to come between them Energy is required to overcome the attrac-tive forces within the solute and within the solvent so both steps are endother-mic When solute and solvent particles mix the particles attract each otherand energy is released This step in the solvation process is exothermic Theoverall energy change that occurs during the solution formation process iscalled the heat of solution

As you observed in the DISCOVERY LAB at the beginning of this chap-ter some solutions release energy as they form whereas others absorb energyduring formation For example after ammonium nitrate dissolves in waterits container feels cool In contrast after calcium chloride dissolves in waterits container feels warm You will learn more about the heat of solution in thenext chapter

SolubilityIf you have ever added so much sugar to a sweetened beverage that sugar crys-tals accumulated on the containerrsquos bottom then you know that only a lim-ited amount of solute can dissolve in a solvent at a given set of conditionsIn fact every solute has a characteristic solubility Solubility refers to the max-imum amount of solute that will dissolve in a given amount of solvent at aspecified temperature and pressure As you can see from Table 15-2 solu-bility is usually expressed in grams of solute per 100 g of solvent

151 What are solutions 457

Solubilities of Some Solutes in Water at Various Temperatures

Substance Formula Solubility (g100 g H2O)

0degC 20degC 60degC 100degC

Aluminum sulfate Al2(SO4)3 312 364 592 890

Ammonium chloride NH4Cl 294 372 553 773

Barium hydroxide Ba(OH)2 167 389 2094 mdash

Barium nitrate Ba(NO3)2 495 902 204 344

Calcium hydroxide Ca(OH)2 0189 0173 0121 0076

Lead(II) chloride PbCl2 067 100 194 320

Lithium sulfate Li2SO4 361 348 326 mdash

Potassium chloride KCl 280 342 458 563

Potassium sulfate K2SO4 74 111 182 241

Sodium chloride NaCl 357 359 371 392

Silver nitrate AgNO3 122 216 440 733

Sucrose C12H22O11 1792 2039 2873 4872

Ammonia NH3 1130 680 200 mdash

Carbon dioxide CO2 1713 0878 0359 mdash

Oxygen O2 0048 0031 0019 mdash

L1 L H2O of gas at standard pressure (101 kPa) mdash No value available

Table 15-2

Earth ScienceCONNECTION

Water is the most abundantsolvent on Earth The

amount of oxygen contained inwater is called dissolved oxygen(DO) Dissolved oxygen is an indi-cator of water quality becauseoxygen is necessary for fish andother aquatic life Oxygen entersa body of water by photosynthe-sis of aquatic plants and by trans-fer of oxygen across the air-waterboundary

Thermal pollution is a reduc-tion in water quality due to anincrease in water temperatureThe discharge of heated waterinto a lake river or other body ofwater by factories or powerplants causes the solubility ofoxygen in water to decrease Inaddition to not having enoughoxygen to support life as shownbelow the lower oxygen levelmagnifies the effects of toxic andorganic pollutants

To reduce thermal pollutionmeasures must be taken beforethe heated water is released intothe environment The water canbe discharged into holding lakesor canals and allowed to cool Inaddition many facilities use cool-ing towers to dissipate heat intothe air

Chem MC-457

null

96913216

Just as solvation can be understood at the par-ticle level so can solubility When a solute isadded to a solvent solvent particles collide withthe solutersquos surface particles solute particles beginto mix randomly among the solvent particles Atfirst the solute particles are carried away from thecrystal However as the number of solvated par-ticles increases the same random mixing results inincreasingly frequent collisions between solvatedsolute particles and the remaining crystal Somecolliding solute particles rejoin the crystal or crys-tallize as you can see in Figure 15-6 As solvationcontinues the crystallization rate increases whilethe solvation rate remains constant As long as thesolvation rate is greater than the crystallizationrate the net effect is continuing solvation

Depending on the amount of solute present the rates of solvation andcrystallization may eventually equalize no more solute appears to dissolveand a state of dynamic equilibrium exists between crystallization and solva-tion (as long as the temperature remains constant) Although solute particlescontinue to dissolve and crystallize in solutions that reach equilibrium theoverall amount of dissolved solute in the solution remains constant Such asolution is said to be a saturated solution it contains the maximum amountof dissolved solute for a given amount of solvent at a specific temperatureand pressure An unsaturated solution is one that contains less dissolvedsolute for a given temperature and pressure than a saturated solution In otherwords more solute can be dissolved in an unsaturated solution

Factors That Affect SolubilityPressure affects the solubility of gaseous solutes and gaseous solutions Thesolubility of a solute also depends on the nature of the solute and solventTemperature affects the solubility of all substances To learn how a blood dis-order called sickle-cell disease can affect oxygenrsquos solubility in blood readthe Chemistry and Society feature at the end of this chapter

Temperature and solubility Many substances aremore soluble at high temperatures than at low tempera-tures as you can see by the data graphed in Figure 15-7For example calcium chloride (CaCl2) has a solubility ofabout 64 g CaCl2 per 100 g H2O at 10degC Increasing thetemperature to approximately 27degC increases the solu-bility by 50 to 100 g CaCl2 per 100 g H2O This factalso is illustrated by the data in Table 15-2 on the previ-ous page From the table you can see that at 20degC 2039 gof sucrose (C12H22O11) dissolves in 100 g of water At100degC 4872 g of sucrose dissolves in 100 g of waternearly a 140 increase in solubility over an 80degC tem-perature range According to Figure 15-7 and Table 15-2what substances decrease in solubility as temperatureincreases

458 Chapter 15 Solutions

0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g o

f so

lute

100

g H

2O) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

CrystallizationSolvation

Figure 15-6

A dynamic equilibrium exists in asaturated solution That is therate at which solute particles inthe crystal are solvated is equalto the rate at which solvatedsolute particles rejoin the crystal

Figure 15-7

This graph shows the solubilityof several substances as a func-tion of temperature What is thesolubility of KClO3 at 60degC

Chem MC-458

null

20219186

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

Just as solvation can be understood at the par-ticle level so can solubility When a solute isadded to a solvent solvent particles collide withthe solutersquos surface particles solute particles beginto mix randomly among the solvent particles Atfirst the solute particles are carried away from thecrystal However as the number of solvated par-ticles increases the same random mixing results inincreasingly frequent collisions between solvatedsolute particles and the remaining crystal Somecolliding solute particles rejoin the crystal or crys-tallize as you can see in Figure 15-6 As solvationcontinues the crystallization rate increases whilethe solvation rate remains constant As long as thesolvation rate is greater than the crystallizationrate the net effect is continuing solvation

Depending on the amount of solute present the rates of solvation andcrystallization may eventually equalize no more solute appears to dissolveand a state of dynamic equilibrium exists between crystallization and solva-tion (as long as the temperature remains constant) Although solute particlescontinue to dissolve and crystallize in solutions that reach equilibrium theoverall amount of dissolved solute in the solution remains constant Such asolution is said to be a saturated solution it contains the maximum amountof dissolved solute for a given amount of solvent at a specific temperatureand pressure An unsaturated solution is one that contains less dissolvedsolute for a given temperature and pressure than a saturated solution In otherwords more solute can be dissolved in an unsaturated solution

Factors That Affect SolubilityPressure affects the solubility of gaseous solutes and gaseous solutions Thesolubility of a solute also depends on the nature of the solute and solventTemperature affects the solubility of all substances To learn how a blood dis-order called sickle-cell disease can affect oxygenrsquos solubility in blood readthe Chemistry and Society feature at the end of this chapter

Temperature and solubility Many substances aremore soluble at high temperatures than at low tempera-tures as you can see by the data graphed in Figure 15-7For example calcium chloride (CaCl2) has a solubility ofabout 64 g CaCl2 per 100 g H2O at 10degC Increasing thetemperature to approximately 27degC increases the solu-bility by 50 to 100 g CaCl2 per 100 g H2O This factalso is illustrated by the data in Table 15-2 on the previ-ous page From the table you can see that at 20degC 2039 gof sucrose (C12H22O11) dissolves in 100 g of water At100degC 4872 g of sucrose dissolves in 100 g of waternearly a 140 increase in solubility over an 80degC tem-perature range According to Figure 15-7 and Table 15-2what substances decrease in solubility as temperatureincreases

458 Chapter 15 Solutions

0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g o

f so

lute

100

g H

2O) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

CrystallizationSolvation

Figure 15-6

A dynamic equilibrium exists in asaturated solution That is therate at which solute particles inthe crystal are solvated is equalto the rate at which solvatedsolute particles rejoin the crystal

Figure 15-7

This graph shows the solubilityof several substances as a func-tion of temperature What is thesolubility of KClO3 at 60degC

Chem MC-458

null

20219186

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

As you can see from Table 15-2 the gases oxygen and carbon dioxide areless soluble at higher temperatures than at lower temperatures This is a pre-dictable trend for all gaseous solutes in liquid solvents Can you explainwhy Recall from Chapter 13 that the kinetic energy of gas particles allowsthem to escape from a solution more readily at higher temperatures Thus asa solutionrsquos temperature increases the solubility of a gaseous solute decreases

The fact that solubility changes with temperature and that some sub-stances become more soluble with increasing temperature is the key toforming supersaturated solutions A supersaturated solution contains moredissolved solute than a saturated solution at the same temperature To makea supersaturated solution a saturated solution is formed at a high tempera-ture and then cooled slowly The slow cooling allows the excess solute toremain dissolved in solution at the lower temperature as shown in Figure15-8a Supersaturated solutions are unstable If a tiny amount of solutecalled a seed crystal is added to a supersaturated solution the excess soluteprecipitates quickly as shown in Figures 15-8b and 15-8c Crystallizationcan also occur if the inside of the container is scratched or the supersatu-rated solution undergoes a physical shock such as stirring or tapping the container Using crystals of silver iodide (AgI) to seed air supersaturated withwater vapor causes the water particles to come together and form dropletsthat may fall to Earth as rain This often-performed technique is called cloudseeding The rock candy and mineral deposits at the edges of mineral springsshown in Figure 15-9 are formed from supersaturated solutions

151 What are solutions 459

Figure 15-9

Examples of crystals that formedfrom supersaturated solutionsinclude rock candy and hotspring mineral deposits

Figure 15-8

Sodium acetate (NaC2H3O2) iscommonly used in the prepara-tion of a supersaturated solu-tion When a seed crystal isadded the excess sodiumacetate quickly crystallizes outof the solution and cb

a

a b c

Chem MC-459

null

11101861

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

Pressure and solubility Pressure affects the solubility of gaseous solutesThe solubility of a gas in any solvent increases as its external pressure (thepressure above the solution) increases Carbonated beverages depend on thisfact Carbonated beverages contain carbon dioxide gas dissolved in an aque-ous solution The dissolved gas gives the beverage its fizz In bottling the bev-erage carbon dioxide is dissolved in the solution at a pressure higher thanatmospheric pressure When the beverage container is opened the pressure ofthe carbon dioxide gas in the space above the liquid (in the neck of the bot-tle) decreases As a result bubbles of carbon dioxide gas form in the solutionrise to the top and escape See Figure 15-10 Unless the cap is placed backon the bottle the process will continue until the solution loses almost all of itscarbon dioxide gas and goes flat

Henryrsquos law The decreased solubility of the carbon dioxide contained in thebeverage after its cap is removed can be described by Henryrsquos law Henryrsquoslaw states that at a given temperature the solubility (S) of a gas in a liquidis directly proportional to the pressure (P) of the gas above the liquid Youcan express this relationship in the following way

PS1

1 P

S2

2

where S1 is the solubility of a gas at a pressure P1 and S2 is the solubility ofthe gas at the new pressure P2

You often will solve Henryrsquos law for the solubility S2 at a new pressure P2where P2 is known The basic rules of algebra can be used to solve Henryrsquoslaw for any one specific variable To solve for S2 begin with the standard formof Henryrsquos law

PS1

1 P

S2

2

Cross-multiplying yields

S1P2 P1S2

Dividing both sides of the equation by P1 yields the desired result the equa-tion solved for S2

SP1P

1

2

PP1S

1

2 S2

SP1P

1

2

460 Chapter 15 Solutions

Figure 15-10

When the cap on the sodabottle is closed pressure abovethe solution keeps excess carbondioxide (CO2) from escaping thesolution When the cap isremoved the decreased pressureabove the solution results in thedecreased solubility of the carbondioxidemdashthe carbon dioxideescapes the solution

b

a

a b

Topic Solutions in the BodyTo learn more about solutionsin the body visit theChemistry Web site atchemistrymccomActivity Research the solu-tions that can be found in thehuman body and in animalsHow do biological systemsmakes use of the principles ofsolubility to support life

151 What are solutions 461

EXAMPLE PROBLEM 15-1

Using Henryrsquos LawIf 085 g of a gas at 40 atm of pressure dissolves in 10 L of water at 25degChow much will dissolve in 10 L of water at 10 atm of pressure and thesame temperature

1 Analyze the ProblemYou are given the solubility of a gas at an initial pressure Thetemperature of the gas remains constant as the pressure changesBecause decreasing pressure reduces a gasrsquos solubility less gas shoulddissolve at the lower pressure

Known Unknown

S1 = 085 gL S2 = gLP1 = 40 atmP2 = 10 atm

2 Solve for the UnknownRearrange Henryrsquos law to solve for S2

PS1

1 =

PS2

2 S2 = S1

PP

2

1

Substitute the known values into the equation and solve

S2 = (085gL)1400

aattmm = 021gL

3 Evaluate the AnswerThe answer is correctly expressed to two significant figures Thesolubility decreased as expected The pressure on the solution was re-duced from 40 atm to 10 atm so the solubility should be reducedto one-fourth its original value which it is

Section 151 Assessment

3 Describe the characteristics of a solution and iden-tify the various types

4 How do intermolecular forces affect solvation

5 What is solubility Describe two factors that affectsolubility

6 Thinking Critically If a seed crystal was added toa supersaturated solution how would you charac-terize the resulting solution

7 Making and Using Graphs Use the information inTable 15-2 to graph the solubilities of aluminumsulfate lithium sulfate and potassium sulfate at0degC 20degC 60degC and 100degC Which substancersquossolubility is most affected by increasing tempera-ture

PRACTICE PROBLEMS1 If 055 g of a gas dissolves in 10 L of water at 200 kPa of pressure

how much will dissolve at 1100 kPa of pressure

2 A gas has a solubility of 066 gL at 100 atm of pressure What is thepressure on a 10-L sample that contains 15 g of gas

For more practice withHenryrsquos law problemsgo to SupplementalPractice Problems

in Appendix A

Practice

chemistrymccomself_check_quiz

462 Chapter 15 Solutions

Section 152 Solution Concentration

Objectivesbull State the concentrations of

solutions in different ways

bull Calculate the concentrationsof solutions

Vocabularyconcentrationmolaritymolalitymole fraction

You have learned about the process of solvation and the factors that affectsolubility The concentration of a solution is a measure of how much soluteis dissolved in a specific amount of solvent or solution How would youdescribe the concentration of the solutions in Figure 15-11 Concentrationmay be described qualitatively using the words concentrated or dilute In gen-eral a concentrated solution as shown on the left in Figure 15-11 containsa large amount of solute Conversely a dilute solution contains a small amountof solute How do you know that the tea on the right in Figure 15-11 is a moredilute solution than the tea on the left

Expressing ConcentrationAlthough qualitative descriptions of concentration can be useful solutions aremore often described quantitatively Some commonly used quantitativedescriptions are percent by either mass or volume molarity and molalityThese descriptions express concentration as a ratio of measured amounts ofsolute and solvent or solution Table 15-3 lists each ratiorsquos description

You may be wondering if onedescription is preferable to anotherThe description used depends onthe type of solution analyzed andthe reason for describing it Forexample a chemist working with areaction in an aqueous solutionmost likely refers to the molarity of the solution because he or sheneeds to know the number of parti-cles involved in the reaction

Figure 15-11

The strength of the tea corre-sponds to its concentration Thedarker cup of tea is more concentrated than the lighter cup

Concentration Ratios

Concentration description Ratio

Percent by mass mmaassss

ooffsosolulutitoen 100

Percent by volume 100

Molarity lmit

oer

leosfosfoslou

ltuioten

Molality

Mole fraction moles of solute

moles of solutekilogram of solvent

volume of solutevolume of solution

Table 15-3

moles of solute moles of solvent

Chem MC-462

null

10177174

CMC_Spn_C15-S02

null

3375004

152 Solution Concentration 463

EXAMPLE PROBLEM 15-2

PRACTICE PROBLEMS8 What is the percent by mass of NaHCO3 in a solution containing 20 g

NaHCO3 dissolved in 600 mL H2O

9 You have 15000 g of a bleach solution The percent by mass of thesolute sodium hypochlorite NaOCl is 362 How many grams ofNaOCl are in the solution

10 In question 9 how many grams of solvent are in the solution

Using Percent to Describe ConcentrationConcentration expressed as a percent is a ratio of a measured amount ofsolute to a measured amount of solution Percent by mass usually describessolutions in which a solid is dissolved in a liquid such as sodium chloride inwater The percent by mass is the ratio of the solutersquos mass to the solutionrsquosmass expressed as a percent The mass of the solution equals the sum of themasses of the solute and the solvent

Percent by mass mmaassss

ooffsosolulutitoen 100

For more practice with percent by mass problems go toSupplemental PracticeSupplemental Practice

Problems in Appendix A

Practice

Maintaining the proper saline(salt) concentration is importantto the health of saltwater fish

Calculating Percent by MassIn order to maintain a sodium chloride (NaCl) concentration similarto ocean water an aquarium must contain 36 g NaCl per 1000 g ofwater What is the percent by mass of NaCl in the solution

1 Analyze the ProblemYou are given the amount of sodium chloride dissolved in 1000 gof water The percent by mass of a solute is the ratio of thesolutersquos mass to the solutionrsquos mass which is the sum of themasses of the solute and the solvent

Known Unknown

mass of solute 36 g NaCl percent by mass mass of solvent 1000 g H2O

2 Solve for the UnknownFind the mass of the solutionMass of solution grams of solute grams of solvent

36 g 1000 g 1036 g

Substitute the known values into the percent by mass equation

Percent by mass mmaassss

ooffsosolulutitoen

100

130366

gg

100

35

3 Evaluate the AnswerBecause only a small mass of sodium chloride is dissolved per 1000 gof water the percent by mass should be a small value which it is Themass of sodium chloride was given with two significant figures there-fore the answer also is expressed with two significant figures

Chem MC-463

null

35787636

464 Chapter 15 Solutions

PRACTICE PROBLEMS11 What is the percent by volume of ethanol in a solution that contains

35 mL of ethanol dissolved in 115 mL of water

12 If you have 1000 mL of a 300 aqueous solution of ethanol whatvolumes of ethanol and water are in the solution

13 What is the percent by volume of isopropyl alcohol in a solution thatcontains 24 mL of isopropyl alcohol in 11 L of water

Percent by volume usually describes solutions in which both solute and sol-vent are liquids The percent by volume is the ratio of the volume of the soluteto the volume of the solution expressed as a percent The volume of the solu-tion is the sum of the volumes of the solute and the solvent Calculations aresimilar to those involving percent by mass

Percent by volume 100

Rubbing alcohol is an aqueous solution of liquid isopropyl alcohol Thelabel on a typical container such as the one shown in Figure 15-12 usuallystates that the rubbing alcohol is 70 isopropyl alcohol This value is a per-cent by volume It tells you that 70 volumes of isopropyl alcohol are dissolvedin every 100 volumes of solution Because a solutionrsquos volume is the sum ofthe volumes of solute and solvent there must be 30 volumes of water (sol-vent) in every 100 volumes of the rubbing alcohol

volume of solutevolume of solution

MolarityAs you have learned percent by volume and percent by mass are only twoof the commonly used ways to quantitatively describe the concentrations ofliquid solutions One of the most common units of solution concentration ismolarity Molarity (M) is the number of moles of solute dissolved per literof solution Molarity also is known as molar concentration The unit M is readas molar A liter of solution containing one mole of solute is a 1M solutionwhich is read as a one molar solution A liter of solution containing 01 moleof solute is a 01M solution

To calculate a solutionrsquos molarity you must know the volume of the solu-tion and the amount of dissolved solute

Molarity (M) lmite

orlsesof

osfosloultuioten

For example suppose you need to calculate the molarity of 1000 mL of anaqueous solution containing 0085 mole of dissolved potassium chloride(KCl) You would first convert the volume of the solution from milliliters toliters using the conversion factor 1 L 1000 mL

(100 mL) 10010LmL 01000 L

Then to determine the molarity you would divide the number of moles ofsolute by the solution volume in liters

001000805

Lmo

sollKut

Cioln

085L

mol 085M

For more practice withpercent by volumeproblems go toSupplemental PracticeSupplemental Practice

Problems in Appendix A

Practice

Figure 15-12

The composition of this isopropylalcohol is given in percent by vol-ume which is often expressed as (vv) What does each v in theexpression 70 (vv) refer to

Chem MC-464

null

16932661

152 Solution Concentration 465

EXAMPLE PROBLEM 15-3

Calculating MolarityA 1005-mL intravenous (IV) solution contains 510 g of glucose(C6H12O6) What is the molarity of this solution The molar mass ofglucose is 18016 gmol

1 Analyze the ProblemYou are given the mass of glucose dissolved in a volume of solu-tion The molarity of the solution is the ratio of moles of soluteper liter of solution Glucose is the solute and water is the solvent

Known Unknown

mass of solute 510 g C6H12O6 solution concentration Mmolar mass of C6H12O6 18016 gmolvolume of solution 1005 mL

2 Solve for the UnknownUse the molar mass to calculate the number of moles of C6H12O6

(510 g C6H12O6) 00283 mol C6H12O6

Use the conversion factor 100

10LmL

to convert the volume of H2Oin milliliters to liters

1005 mL solution 10010LmL

01005 L solution

Substitute the known values into the equation for molarity and solve

molarity lmite

orlses

ofof

sosloultuiote

n

molarity 0282M

3 Evaluate the AnswerThe molarity is a small value which is expected because only a smallmass of glucose was dissolved in the solution The mass of glucoseused in the problem contained three significant figures and there-fore the value of the molarity also has three significant figures

0282 mol C6H12O6

L solution00283 mol C6H12O6

01005 L solution

PRACTICE PROBLEMS14 What is the molarity of an aqueous solution containing 400 g of

glucose (C6H12O6) in 15 L of solution

15 What is the molarity of a bleach solution containing 95 g of NaOClper liter of bleach

16 Calculate the molarity of 160 L of a solution containing 155 g of dissolved KBr

For more practice withmolarity problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

To prevent dehydration intra-venous (IV) drips are administeredto many hospital patients A solution containing sodium chlo-ride and glucose is commonlyused

Do the CHEMLAB at the end of this chapter to learn about an experi-mental technique for determining solution concentration

1 mol C6H12O6

18016 g C6H12O6

Chem MC-465

null

6373885

Preparing Molar Solutions Now that you know how to calculate the molarity of a solution how do youthink you would prepare 1 L of a 150M aqueous solution of sucrose(C12H22O11) for an experiment A 150M aqueous solution of sucrose con-tains 150 moles of sucrose dissolved in a liter of solution The molar massof sucrose is 342 g Thus 150 moles of sucrose has a mass of 513 g anamount that you can measure on a balance

150

1m

Lol

soCl1

u2

tHio

2

n2O11

314m2

oglCC

1

1

2

2

HH

2

2

2

2

OO

1

1

1

1

5131

gL

Cso

12

luHt2

io2On

11

Unfortunately you cannot simply add 513 g of sugar to one liter of waterto make the 150M solution Do you know why Like all substances sugartakes up space and will add volume to the solution Therefore you must useslightly less than one liter of water to make one liter of solution Follow thesteps shown in Figure 15-13 to learn how to prepare the correct volume ofthe solution

You often will do experiments that call for only small quantities of solu-tion For example you may need only 100 mL of a 150M sucrose solutionfor an experiment How do you determine the amount of sucrose to use Lookagain at the definition of molarity As calculated above 150M solution ofsucrose contains 150 mol of sucrose per one liter of solution Therefore oneliter of solution contains 513 g of sucrose

This relationship can be used as a conversion factor to calculate how muchsolute you need for your experiment

100 mL 10010LmL

5113

Lg C

so1

l2

uHti

2

o2

nO11

513 g C12H22O11

Thus you would need to measure out 513 g of sucrose to make 100 mL ofa 150M solution

466 Chapter 15 Solutions

Figure 15-13

Accurately preparing a solutiontakes care In step 1 the massof solute to be used is measuredout In step 2 the solute isplaced in a volumetric flask ofthe correct volume In step 3distilled water is added to theflask to bring the solution levelup to the calibration mark onthe flask

c

b

a

For more practice withmolarity problems goto SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

PRACTICE PROBLEMS17 How many grams of CaCl2 would be dissolved in 10 L of a 010M

solution of CaCl2

18 A liter of 2M NaOH solution contains how many grams of NaOH

19 How many grams of CaCl2 should be dissolved in 5000 mL of waterto make a 020M solution of CaCl2

20 How many grams of NaOH are in 250 mL of a 30M NaOH solution

a b c

Chem MC-466

null

14335835

Diluting solutions In the laboratory you may use concentrated solutions ofstandard molarities called stock solutions For example concentrated hydrochlo-ric acid (HCl) is 12M Recall that a concentrated solution has a large amountof solute You can prepare a less concentrated solution by diluting the stock solu-tion with solvent When you add solvent you increase the number of solventparticles among which the solute particles move as shown in Figure 15-14thereby decreasing the solutionrsquos concentration Would you still have the samenumber of moles of solute particles that were in the stock solution Why

How do you determine the volume of stock solution you must dilute Youknow that

Molarity (M) lmite

orlsesof

osfosloultuioten

You can rearrange the expression of molarity to solve for moles of solute

Moles of solute molarity liters of solution

Because the total number of moles of solute does not change during dilution

Moles of solute in the stock solution moles of solute after dilution

You can write this relationship as the expression

where M1 and V1 represent the molarity and volume of the stock solution andM2 and V2 represent the molarity and volume of the dilute solution

M1V1 M2V2

152 Solution Concentration 467

Figure 15-14

The concentration of a solutioncan be diluted by adding addi-tional solvent

After adding additional solvent to the solutionthe ratio of solute particles to solvent particles(water molecules) has decreased This solution isless concentrated than the solution in a

b

Before dilution this solutioncontains a fairly high ratio ofsolute particles to solvent particles(water molecules)

a

Dilute solutionM2V2

Concentrated solutionM1V1

Add solvent

SoluteSolvent (water)

Chem MC-467

null

82964554

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

152 Solution Concentration 463

EXAMPLE PROBLEM 15-2

PRACTICE PROBLEMS8 What is the percent by mass of NaHCO3 in a solution containing 20 g

NaHCO3 dissolved in 600 mL H2O

9 You have 15000 g of a bleach solution The percent by mass of thesolute sodium hypochlorite NaOCl is 362 How many grams ofNaOCl are in the solution

10 In question 9 how many grams of solvent are in the solution

Using Percent to Describe ConcentrationConcentration expressed as a percent is a ratio of a measured amount ofsolute to a measured amount of solution Percent by mass usually describessolutions in which a solid is dissolved in a liquid such as sodium chloride inwater The percent by mass is the ratio of the solutersquos mass to the solutionrsquosmass expressed as a percent The mass of the solution equals the sum of themasses of the solute and the solvent

Percent by mass mmaassss

ooffsosolulutitoen 100

For more practice with percent by mass problems go toSupplemental PracticeSupplemental Practice

Problems in Appendix A

Practice

Maintaining the proper saline(salt) concentration is importantto the health of saltwater fish

Calculating Percent by MassIn order to maintain a sodium chloride (NaCl) concentration similarto ocean water an aquarium must contain 36 g NaCl per 1000 g ofwater What is the percent by mass of NaCl in the solution

1 Analyze the ProblemYou are given the amount of sodium chloride dissolved in 1000 gof water The percent by mass of a solute is the ratio of thesolutersquos mass to the solutionrsquos mass which is the sum of themasses of the solute and the solvent

Known Unknown

mass of solute 36 g NaCl percent by mass mass of solvent 1000 g H2O

2 Solve for the UnknownFind the mass of the solutionMass of solution grams of solute grams of solvent

36 g 1000 g 1036 g

Substitute the known values into the percent by mass equation

Percent by mass mmaassss

ooffsosolulutitoen

100

130366

gg

100

35

3 Evaluate the AnswerBecause only a small mass of sodium chloride is dissolved per 1000 gof water the percent by mass should be a small value which it is Themass of sodium chloride was given with two significant figures there-fore the answer also is expressed with two significant figures

Chem MC-463

null

35787636

464 Chapter 15 Solutions

PRACTICE PROBLEMS11 What is the percent by volume of ethanol in a solution that contains

35 mL of ethanol dissolved in 115 mL of water

12 If you have 1000 mL of a 300 aqueous solution of ethanol whatvolumes of ethanol and water are in the solution

13 What is the percent by volume of isopropyl alcohol in a solution thatcontains 24 mL of isopropyl alcohol in 11 L of water

Percent by volume usually describes solutions in which both solute and sol-vent are liquids The percent by volume is the ratio of the volume of the soluteto the volume of the solution expressed as a percent The volume of the solu-tion is the sum of the volumes of the solute and the solvent Calculations aresimilar to those involving percent by mass

Percent by volume 100

Rubbing alcohol is an aqueous solution of liquid isopropyl alcohol Thelabel on a typical container such as the one shown in Figure 15-12 usuallystates that the rubbing alcohol is 70 isopropyl alcohol This value is a per-cent by volume It tells you that 70 volumes of isopropyl alcohol are dissolvedin every 100 volumes of solution Because a solutionrsquos volume is the sum ofthe volumes of solute and solvent there must be 30 volumes of water (sol-vent) in every 100 volumes of the rubbing alcohol

volume of solutevolume of solution

MolarityAs you have learned percent by volume and percent by mass are only twoof the commonly used ways to quantitatively describe the concentrations ofliquid solutions One of the most common units of solution concentration ismolarity Molarity (M) is the number of moles of solute dissolved per literof solution Molarity also is known as molar concentration The unit M is readas molar A liter of solution containing one mole of solute is a 1M solutionwhich is read as a one molar solution A liter of solution containing 01 moleof solute is a 01M solution

To calculate a solutionrsquos molarity you must know the volume of the solu-tion and the amount of dissolved solute

Molarity (M) lmite

orlsesof

osfosloultuioten

For example suppose you need to calculate the molarity of 1000 mL of anaqueous solution containing 0085 mole of dissolved potassium chloride(KCl) You would first convert the volume of the solution from milliliters toliters using the conversion factor 1 L 1000 mL

(100 mL) 10010LmL 01000 L

Then to determine the molarity you would divide the number of moles ofsolute by the solution volume in liters

001000805

Lmo

sollKut

Cioln

085L

mol 085M

For more practice withpercent by volumeproblems go toSupplemental PracticeSupplemental Practice

Problems in Appendix A

Practice

Figure 15-12

The composition of this isopropylalcohol is given in percent by vol-ume which is often expressed as (vv) What does each v in theexpression 70 (vv) refer to

Chem MC-464

null

16932661

152 Solution Concentration 465

EXAMPLE PROBLEM 15-3

Calculating MolarityA 1005-mL intravenous (IV) solution contains 510 g of glucose(C6H12O6) What is the molarity of this solution The molar mass ofglucose is 18016 gmol

1 Analyze the ProblemYou are given the mass of glucose dissolved in a volume of solu-tion The molarity of the solution is the ratio of moles of soluteper liter of solution Glucose is the solute and water is the solvent

Known Unknown

mass of solute 510 g C6H12O6 solution concentration Mmolar mass of C6H12O6 18016 gmolvolume of solution 1005 mL

2 Solve for the UnknownUse the molar mass to calculate the number of moles of C6H12O6

(510 g C6H12O6) 00283 mol C6H12O6

Use the conversion factor 100

10LmL

to convert the volume of H2Oin milliliters to liters

1005 mL solution 10010LmL

01005 L solution

Substitute the known values into the equation for molarity and solve

molarity lmite

orlses

ofof

sosloultuiote

n

molarity 0282M

3 Evaluate the AnswerThe molarity is a small value which is expected because only a smallmass of glucose was dissolved in the solution The mass of glucoseused in the problem contained three significant figures and there-fore the value of the molarity also has three significant figures

0282 mol C6H12O6

L solution00283 mol C6H12O6

01005 L solution

PRACTICE PROBLEMS14 What is the molarity of an aqueous solution containing 400 g of

glucose (C6H12O6) in 15 L of solution

15 What is the molarity of a bleach solution containing 95 g of NaOClper liter of bleach

16 Calculate the molarity of 160 L of a solution containing 155 g of dissolved KBr

For more practice withmolarity problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

To prevent dehydration intra-venous (IV) drips are administeredto many hospital patients A solution containing sodium chlo-ride and glucose is commonlyused

Do the CHEMLAB at the end of this chapter to learn about an experi-mental technique for determining solution concentration

1 mol C6H12O6

18016 g C6H12O6

Chem MC-465

null

6373885

Preparing Molar Solutions Now that you know how to calculate the molarity of a solution how do youthink you would prepare 1 L of a 150M aqueous solution of sucrose(C12H22O11) for an experiment A 150M aqueous solution of sucrose con-tains 150 moles of sucrose dissolved in a liter of solution The molar massof sucrose is 342 g Thus 150 moles of sucrose has a mass of 513 g anamount that you can measure on a balance

150

1m

Lol

soCl1

u2

tHio

2

n2O11

314m2

oglCC

1

1

2

2

HH

2

2

2

2

OO

1

1

1

1

5131

gL

Cso

12

luHt2

io2On

11

Unfortunately you cannot simply add 513 g of sugar to one liter of waterto make the 150M solution Do you know why Like all substances sugartakes up space and will add volume to the solution Therefore you must useslightly less than one liter of water to make one liter of solution Follow thesteps shown in Figure 15-13 to learn how to prepare the correct volume ofthe solution

You often will do experiments that call for only small quantities of solu-tion For example you may need only 100 mL of a 150M sucrose solutionfor an experiment How do you determine the amount of sucrose to use Lookagain at the definition of molarity As calculated above 150M solution ofsucrose contains 150 mol of sucrose per one liter of solution Therefore oneliter of solution contains 513 g of sucrose

This relationship can be used as a conversion factor to calculate how muchsolute you need for your experiment

100 mL 10010LmL

5113

Lg C

so1

l2

uHti

2

o2

nO11

513 g C12H22O11

Thus you would need to measure out 513 g of sucrose to make 100 mL ofa 150M solution

466 Chapter 15 Solutions

Figure 15-13

Accurately preparing a solutiontakes care In step 1 the massof solute to be used is measuredout In step 2 the solute isplaced in a volumetric flask ofthe correct volume In step 3distilled water is added to theflask to bring the solution levelup to the calibration mark onthe flask

c

b

a

For more practice withmolarity problems goto SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

PRACTICE PROBLEMS17 How many grams of CaCl2 would be dissolved in 10 L of a 010M

solution of CaCl2

18 A liter of 2M NaOH solution contains how many grams of NaOH

19 How many grams of CaCl2 should be dissolved in 5000 mL of waterto make a 020M solution of CaCl2

20 How many grams of NaOH are in 250 mL of a 30M NaOH solution

a b c

Chem MC-466

null

14335835

Diluting solutions In the laboratory you may use concentrated solutions ofstandard molarities called stock solutions For example concentrated hydrochlo-ric acid (HCl) is 12M Recall that a concentrated solution has a large amountof solute You can prepare a less concentrated solution by diluting the stock solu-tion with solvent When you add solvent you increase the number of solventparticles among which the solute particles move as shown in Figure 15-14thereby decreasing the solutionrsquos concentration Would you still have the samenumber of moles of solute particles that were in the stock solution Why

How do you determine the volume of stock solution you must dilute Youknow that

Molarity (M) lmite

orlsesof

osfosloultuioten

You can rearrange the expression of molarity to solve for moles of solute

Moles of solute molarity liters of solution

Because the total number of moles of solute does not change during dilution

Moles of solute in the stock solution moles of solute after dilution

You can write this relationship as the expression

where M1 and V1 represent the molarity and volume of the stock solution andM2 and V2 represent the molarity and volume of the dilute solution

M1V1 M2V2

152 Solution Concentration 467

Figure 15-14

The concentration of a solutioncan be diluted by adding addi-tional solvent

After adding additional solvent to the solutionthe ratio of solute particles to solvent particles(water molecules) has decreased This solution isless concentrated than the solution in a

b

Before dilution this solutioncontains a fairly high ratio ofsolute particles to solvent particles(water molecules)

a

Dilute solutionM2V2

Concentrated solutionM1V1

Add solvent

SoluteSolvent (water)

Chem MC-467

null

82964554

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

464 Chapter 15 Solutions

PRACTICE PROBLEMS11 What is the percent by volume of ethanol in a solution that contains

35 mL of ethanol dissolved in 115 mL of water

12 If you have 1000 mL of a 300 aqueous solution of ethanol whatvolumes of ethanol and water are in the solution

13 What is the percent by volume of isopropyl alcohol in a solution thatcontains 24 mL of isopropyl alcohol in 11 L of water

Percent by volume usually describes solutions in which both solute and sol-vent are liquids The percent by volume is the ratio of the volume of the soluteto the volume of the solution expressed as a percent The volume of the solu-tion is the sum of the volumes of the solute and the solvent Calculations aresimilar to those involving percent by mass

Percent by volume 100

Rubbing alcohol is an aqueous solution of liquid isopropyl alcohol Thelabel on a typical container such as the one shown in Figure 15-12 usuallystates that the rubbing alcohol is 70 isopropyl alcohol This value is a per-cent by volume It tells you that 70 volumes of isopropyl alcohol are dissolvedin every 100 volumes of solution Because a solutionrsquos volume is the sum ofthe volumes of solute and solvent there must be 30 volumes of water (sol-vent) in every 100 volumes of the rubbing alcohol

volume of solutevolume of solution

MolarityAs you have learned percent by volume and percent by mass are only twoof the commonly used ways to quantitatively describe the concentrations ofliquid solutions One of the most common units of solution concentration ismolarity Molarity (M) is the number of moles of solute dissolved per literof solution Molarity also is known as molar concentration The unit M is readas molar A liter of solution containing one mole of solute is a 1M solutionwhich is read as a one molar solution A liter of solution containing 01 moleof solute is a 01M solution

To calculate a solutionrsquos molarity you must know the volume of the solu-tion and the amount of dissolved solute

Molarity (M) lmite

orlsesof

osfosloultuioten

For example suppose you need to calculate the molarity of 1000 mL of anaqueous solution containing 0085 mole of dissolved potassium chloride(KCl) You would first convert the volume of the solution from milliliters toliters using the conversion factor 1 L 1000 mL

(100 mL) 10010LmL 01000 L

Then to determine the molarity you would divide the number of moles ofsolute by the solution volume in liters

001000805

Lmo

sollKut

Cioln

085L

mol 085M

For more practice withpercent by volumeproblems go toSupplemental PracticeSupplemental Practice

Problems in Appendix A

Practice

Figure 15-12

The composition of this isopropylalcohol is given in percent by vol-ume which is often expressed as (vv) What does each v in theexpression 70 (vv) refer to

Chem MC-464

null

16932661

152 Solution Concentration 465

EXAMPLE PROBLEM 15-3

Calculating MolarityA 1005-mL intravenous (IV) solution contains 510 g of glucose(C6H12O6) What is the molarity of this solution The molar mass ofglucose is 18016 gmol

1 Analyze the ProblemYou are given the mass of glucose dissolved in a volume of solu-tion The molarity of the solution is the ratio of moles of soluteper liter of solution Glucose is the solute and water is the solvent

Known Unknown

mass of solute 510 g C6H12O6 solution concentration Mmolar mass of C6H12O6 18016 gmolvolume of solution 1005 mL

2 Solve for the UnknownUse the molar mass to calculate the number of moles of C6H12O6

(510 g C6H12O6) 00283 mol C6H12O6

Use the conversion factor 100

10LmL

to convert the volume of H2Oin milliliters to liters

1005 mL solution 10010LmL

01005 L solution

Substitute the known values into the equation for molarity and solve

molarity lmite

orlses

ofof

sosloultuiote

n

molarity 0282M

3 Evaluate the AnswerThe molarity is a small value which is expected because only a smallmass of glucose was dissolved in the solution The mass of glucoseused in the problem contained three significant figures and there-fore the value of the molarity also has three significant figures

0282 mol C6H12O6

L solution00283 mol C6H12O6

01005 L solution

PRACTICE PROBLEMS14 What is the molarity of an aqueous solution containing 400 g of

glucose (C6H12O6) in 15 L of solution

15 What is the molarity of a bleach solution containing 95 g of NaOClper liter of bleach

16 Calculate the molarity of 160 L of a solution containing 155 g of dissolved KBr

For more practice withmolarity problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

To prevent dehydration intra-venous (IV) drips are administeredto many hospital patients A solution containing sodium chlo-ride and glucose is commonlyused

Do the CHEMLAB at the end of this chapter to learn about an experi-mental technique for determining solution concentration

1 mol C6H12O6

18016 g C6H12O6

Chem MC-465

null

6373885

Preparing Molar Solutions Now that you know how to calculate the molarity of a solution how do youthink you would prepare 1 L of a 150M aqueous solution of sucrose(C12H22O11) for an experiment A 150M aqueous solution of sucrose con-tains 150 moles of sucrose dissolved in a liter of solution The molar massof sucrose is 342 g Thus 150 moles of sucrose has a mass of 513 g anamount that you can measure on a balance

150

1m

Lol

soCl1

u2

tHio

2

n2O11

314m2

oglCC

1

1

2

2

HH

2

2

2

2

OO

1

1

1

1

5131

gL

Cso

12

luHt2

io2On

11

Unfortunately you cannot simply add 513 g of sugar to one liter of waterto make the 150M solution Do you know why Like all substances sugartakes up space and will add volume to the solution Therefore you must useslightly less than one liter of water to make one liter of solution Follow thesteps shown in Figure 15-13 to learn how to prepare the correct volume ofthe solution

You often will do experiments that call for only small quantities of solu-tion For example you may need only 100 mL of a 150M sucrose solutionfor an experiment How do you determine the amount of sucrose to use Lookagain at the definition of molarity As calculated above 150M solution ofsucrose contains 150 mol of sucrose per one liter of solution Therefore oneliter of solution contains 513 g of sucrose

This relationship can be used as a conversion factor to calculate how muchsolute you need for your experiment

100 mL 10010LmL

5113

Lg C

so1

l2

uHti

2

o2

nO11

513 g C12H22O11

Thus you would need to measure out 513 g of sucrose to make 100 mL ofa 150M solution

466 Chapter 15 Solutions

Figure 15-13

Accurately preparing a solutiontakes care In step 1 the massof solute to be used is measuredout In step 2 the solute isplaced in a volumetric flask ofthe correct volume In step 3distilled water is added to theflask to bring the solution levelup to the calibration mark onthe flask

c

b

a

For more practice withmolarity problems goto SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

PRACTICE PROBLEMS17 How many grams of CaCl2 would be dissolved in 10 L of a 010M

solution of CaCl2

18 A liter of 2M NaOH solution contains how many grams of NaOH

19 How many grams of CaCl2 should be dissolved in 5000 mL of waterto make a 020M solution of CaCl2

20 How many grams of NaOH are in 250 mL of a 30M NaOH solution

a b c

Chem MC-466

null

14335835

Diluting solutions In the laboratory you may use concentrated solutions ofstandard molarities called stock solutions For example concentrated hydrochlo-ric acid (HCl) is 12M Recall that a concentrated solution has a large amountof solute You can prepare a less concentrated solution by diluting the stock solu-tion with solvent When you add solvent you increase the number of solventparticles among which the solute particles move as shown in Figure 15-14thereby decreasing the solutionrsquos concentration Would you still have the samenumber of moles of solute particles that were in the stock solution Why

How do you determine the volume of stock solution you must dilute Youknow that

Molarity (M) lmite

orlsesof

osfosloultuioten

You can rearrange the expression of molarity to solve for moles of solute

Moles of solute molarity liters of solution

Because the total number of moles of solute does not change during dilution

Moles of solute in the stock solution moles of solute after dilution

You can write this relationship as the expression

where M1 and V1 represent the molarity and volume of the stock solution andM2 and V2 represent the molarity and volume of the dilute solution

M1V1 M2V2

152 Solution Concentration 467

Figure 15-14

The concentration of a solutioncan be diluted by adding addi-tional solvent

After adding additional solvent to the solutionthe ratio of solute particles to solvent particles(water molecules) has decreased This solution isless concentrated than the solution in a

b

Before dilution this solutioncontains a fairly high ratio ofsolute particles to solvent particles(water molecules)

a

Dilute solutionM2V2

Concentrated solutionM1V1

Add solvent

SoluteSolvent (water)

Chem MC-467

null

82964554

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

152 Solution Concentration 465

EXAMPLE PROBLEM 15-3

Calculating MolarityA 1005-mL intravenous (IV) solution contains 510 g of glucose(C6H12O6) What is the molarity of this solution The molar mass ofglucose is 18016 gmol

1 Analyze the ProblemYou are given the mass of glucose dissolved in a volume of solu-tion The molarity of the solution is the ratio of moles of soluteper liter of solution Glucose is the solute and water is the solvent

Known Unknown

mass of solute 510 g C6H12O6 solution concentration Mmolar mass of C6H12O6 18016 gmolvolume of solution 1005 mL

2 Solve for the UnknownUse the molar mass to calculate the number of moles of C6H12O6

(510 g C6H12O6) 00283 mol C6H12O6

Use the conversion factor 100

10LmL

to convert the volume of H2Oin milliliters to liters

1005 mL solution 10010LmL

01005 L solution

Substitute the known values into the equation for molarity and solve

molarity lmite

orlses

ofof

sosloultuiote

n

molarity 0282M

3 Evaluate the AnswerThe molarity is a small value which is expected because only a smallmass of glucose was dissolved in the solution The mass of glucoseused in the problem contained three significant figures and there-fore the value of the molarity also has three significant figures

0282 mol C6H12O6

L solution00283 mol C6H12O6

01005 L solution

PRACTICE PROBLEMS14 What is the molarity of an aqueous solution containing 400 g of

glucose (C6H12O6) in 15 L of solution

15 What is the molarity of a bleach solution containing 95 g of NaOClper liter of bleach

16 Calculate the molarity of 160 L of a solution containing 155 g of dissolved KBr

For more practice withmolarity problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

To prevent dehydration intra-venous (IV) drips are administeredto many hospital patients A solution containing sodium chlo-ride and glucose is commonlyused

Do the CHEMLAB at the end of this chapter to learn about an experi-mental technique for determining solution concentration

1 mol C6H12O6

18016 g C6H12O6

Chem MC-465

null

6373885

Preparing Molar Solutions Now that you know how to calculate the molarity of a solution how do youthink you would prepare 1 L of a 150M aqueous solution of sucrose(C12H22O11) for an experiment A 150M aqueous solution of sucrose con-tains 150 moles of sucrose dissolved in a liter of solution The molar massof sucrose is 342 g Thus 150 moles of sucrose has a mass of 513 g anamount that you can measure on a balance

150

1m

Lol

soCl1

u2

tHio

2

n2O11

314m2

oglCC

1

1

2

2

HH

2

2

2

2

OO

1

1

1

1

5131

gL

Cso

12

luHt2

io2On

11

Unfortunately you cannot simply add 513 g of sugar to one liter of waterto make the 150M solution Do you know why Like all substances sugartakes up space and will add volume to the solution Therefore you must useslightly less than one liter of water to make one liter of solution Follow thesteps shown in Figure 15-13 to learn how to prepare the correct volume ofthe solution

You often will do experiments that call for only small quantities of solu-tion For example you may need only 100 mL of a 150M sucrose solutionfor an experiment How do you determine the amount of sucrose to use Lookagain at the definition of molarity As calculated above 150M solution ofsucrose contains 150 mol of sucrose per one liter of solution Therefore oneliter of solution contains 513 g of sucrose

This relationship can be used as a conversion factor to calculate how muchsolute you need for your experiment

100 mL 10010LmL

5113

Lg C

so1

l2

uHti

2

o2

nO11

513 g C12H22O11

Thus you would need to measure out 513 g of sucrose to make 100 mL ofa 150M solution

466 Chapter 15 Solutions

Figure 15-13

Accurately preparing a solutiontakes care In step 1 the massof solute to be used is measuredout In step 2 the solute isplaced in a volumetric flask ofthe correct volume In step 3distilled water is added to theflask to bring the solution levelup to the calibration mark onthe flask

c

b

a

For more practice withmolarity problems goto SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

PRACTICE PROBLEMS17 How many grams of CaCl2 would be dissolved in 10 L of a 010M

solution of CaCl2

18 A liter of 2M NaOH solution contains how many grams of NaOH

19 How many grams of CaCl2 should be dissolved in 5000 mL of waterto make a 020M solution of CaCl2

20 How many grams of NaOH are in 250 mL of a 30M NaOH solution

a b c

Chem MC-466

null

14335835

Diluting solutions In the laboratory you may use concentrated solutions ofstandard molarities called stock solutions For example concentrated hydrochlo-ric acid (HCl) is 12M Recall that a concentrated solution has a large amountof solute You can prepare a less concentrated solution by diluting the stock solu-tion with solvent When you add solvent you increase the number of solventparticles among which the solute particles move as shown in Figure 15-14thereby decreasing the solutionrsquos concentration Would you still have the samenumber of moles of solute particles that were in the stock solution Why

How do you determine the volume of stock solution you must dilute Youknow that

Molarity (M) lmite

orlsesof

osfosloultuioten

You can rearrange the expression of molarity to solve for moles of solute

Moles of solute molarity liters of solution

Because the total number of moles of solute does not change during dilution

Moles of solute in the stock solution moles of solute after dilution

You can write this relationship as the expression

where M1 and V1 represent the molarity and volume of the stock solution andM2 and V2 represent the molarity and volume of the dilute solution

M1V1 M2V2

152 Solution Concentration 467

Figure 15-14

The concentration of a solutioncan be diluted by adding addi-tional solvent

After adding additional solvent to the solutionthe ratio of solute particles to solvent particles(water molecules) has decreased This solution isless concentrated than the solution in a

b

Before dilution this solutioncontains a fairly high ratio ofsolute particles to solvent particles(water molecules)

a

Dilute solutionM2V2

Concentrated solutionM1V1

Add solvent

SoluteSolvent (water)

Chem MC-467

null

82964554

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

Preparing Molar Solutions Now that you know how to calculate the molarity of a solution how do youthink you would prepare 1 L of a 150M aqueous solution of sucrose(C12H22O11) for an experiment A 150M aqueous solution of sucrose con-tains 150 moles of sucrose dissolved in a liter of solution The molar massof sucrose is 342 g Thus 150 moles of sucrose has a mass of 513 g anamount that you can measure on a balance

150

1m

Lol

soCl1

u2

tHio

2

n2O11

314m2

oglCC

1

1

2

2

HH

2

2

2

2

OO

1

1

1

1

5131

gL

Cso

12

luHt2

io2On

11

Unfortunately you cannot simply add 513 g of sugar to one liter of waterto make the 150M solution Do you know why Like all substances sugartakes up space and will add volume to the solution Therefore you must useslightly less than one liter of water to make one liter of solution Follow thesteps shown in Figure 15-13 to learn how to prepare the correct volume ofthe solution

You often will do experiments that call for only small quantities of solu-tion For example you may need only 100 mL of a 150M sucrose solutionfor an experiment How do you determine the amount of sucrose to use Lookagain at the definition of molarity As calculated above 150M solution ofsucrose contains 150 mol of sucrose per one liter of solution Therefore oneliter of solution contains 513 g of sucrose

This relationship can be used as a conversion factor to calculate how muchsolute you need for your experiment

100 mL 10010LmL

5113

Lg C

so1

l2

uHti

2

o2

nO11

513 g C12H22O11

Thus you would need to measure out 513 g of sucrose to make 100 mL ofa 150M solution

466 Chapter 15 Solutions

Figure 15-13

Accurately preparing a solutiontakes care In step 1 the massof solute to be used is measuredout In step 2 the solute isplaced in a volumetric flask ofthe correct volume In step 3distilled water is added to theflask to bring the solution levelup to the calibration mark onthe flask

c

b

a

For more practice withmolarity problems goto SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

PRACTICE PROBLEMS17 How many grams of CaCl2 would be dissolved in 10 L of a 010M

solution of CaCl2

18 A liter of 2M NaOH solution contains how many grams of NaOH

19 How many grams of CaCl2 should be dissolved in 5000 mL of waterto make a 020M solution of CaCl2

20 How many grams of NaOH are in 250 mL of a 30M NaOH solution

a b c

Chem MC-466

null

14335835

Diluting solutions In the laboratory you may use concentrated solutions ofstandard molarities called stock solutions For example concentrated hydrochlo-ric acid (HCl) is 12M Recall that a concentrated solution has a large amountof solute You can prepare a less concentrated solution by diluting the stock solu-tion with solvent When you add solvent you increase the number of solventparticles among which the solute particles move as shown in Figure 15-14thereby decreasing the solutionrsquos concentration Would you still have the samenumber of moles of solute particles that were in the stock solution Why

How do you determine the volume of stock solution you must dilute Youknow that

Molarity (M) lmite

orlsesof

osfosloultuioten

You can rearrange the expression of molarity to solve for moles of solute

Moles of solute molarity liters of solution

Because the total number of moles of solute does not change during dilution

Moles of solute in the stock solution moles of solute after dilution

You can write this relationship as the expression

where M1 and V1 represent the molarity and volume of the stock solution andM2 and V2 represent the molarity and volume of the dilute solution

M1V1 M2V2

152 Solution Concentration 467

Figure 15-14

The concentration of a solutioncan be diluted by adding addi-tional solvent

After adding additional solvent to the solutionthe ratio of solute particles to solvent particles(water molecules) has decreased This solution isless concentrated than the solution in a

b

Before dilution this solutioncontains a fairly high ratio ofsolute particles to solvent particles(water molecules)

a

Dilute solutionM2V2

Concentrated solutionM1V1

Add solvent

SoluteSolvent (water)

Chem MC-467

null

82964554

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

Diluting solutions In the laboratory you may use concentrated solutions ofstandard molarities called stock solutions For example concentrated hydrochlo-ric acid (HCl) is 12M Recall that a concentrated solution has a large amountof solute You can prepare a less concentrated solution by diluting the stock solu-tion with solvent When you add solvent you increase the number of solventparticles among which the solute particles move as shown in Figure 15-14thereby decreasing the solutionrsquos concentration Would you still have the samenumber of moles of solute particles that were in the stock solution Why

How do you determine the volume of stock solution you must dilute Youknow that

Molarity (M) lmite

orlsesof

osfosloultuioten

You can rearrange the expression of molarity to solve for moles of solute

Moles of solute molarity liters of solution

Because the total number of moles of solute does not change during dilution

Moles of solute in the stock solution moles of solute after dilution

You can write this relationship as the expression

where M1 and V1 represent the molarity and volume of the stock solution andM2 and V2 represent the molarity and volume of the dilute solution

M1V1 M2V2

152 Solution Concentration 467

Figure 15-14

The concentration of a solutioncan be diluted by adding addi-tional solvent

After adding additional solvent to the solutionthe ratio of solute particles to solvent particles(water molecules) has decreased This solution isless concentrated than the solution in a

b

Before dilution this solutioncontains a fairly high ratio ofsolute particles to solvent particles(water molecules)

a

Dilute solutionM2V2

Concentrated solutionM1V1

Add solvent

SoluteSolvent (water)

Chem MC-467

null

82964554

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

468 Chapter 15 Solutions

EXAMPLE PROBLEM 15-4

Diluting Stock SolutionsWhat volume in milliliters of 200M calcium chloride (CaCl2) stock solution would you use to make 050 L of 0300M calcium chloridesolution

1 Analyze the ProblemYou are given the molarity of a stock solution of CaCl2 and the vol-ume and molarity of a dilute solution of CaCl2 Use the relationshipbetween molarities and volumes to find the volume in liters of thestock solution required Then convert the volume to milliliters

Known Unknown

M1 200M CaCl2 V1 L 200M CaCl2M2 0300MV2 050 L

2 Solve for the UnknownSolve the molarityndashvolume relationship for the volume of the stocksolution V1

M1V1 M2V2

Dividing both sides of the equation yields

V1 V2 MM

2

1

Substitute the known values into the equation and solve

V1 (050 L)02300000MM

V1 0075 L

Use the conversion factor 10010LmL

to convert the volume from litersto milliliters

V1 (0075 L)10010LmL

75 mL

To make the dilution measure out 75 mL of the stock solution anddilute it with enough water to make the final volume 050 L

3 Evaluate the AnswerThe volume V1 was calculated and then its value was converted tomilliliters Of the given information V2 had the fewest number of sig-nificant figures with two Thus the volume V1 should also have twosignificant figures as it does

PRACTICE PROBLEMS21 What volume of a 300M KI stock solution would you use to make

0300 L of a 125M KI solution

22 How many milliliters of a 50M H2SO4 stock solution would you needto prepare 1000 mL of 025M H2SO4

23 If you dilute 200 mL of a 35M solution to make 1000 mL of solu-tion what is the molarity of the dilute solution

For more practice withdilution problems go to SupplementalSupplementalPractice ProblemsPractice Problems in

Appendix A

Practice

Knowing how to apply the equa-tion M1V1 M2V2 makes thepreparation of dilute solutionseasy

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

152 Solution Concentration 469

Molality and Mole FractionThe volume of a solution changes with temperature as it expands or contractsThis change in volume alters the molarity of the solution Masses howeverdo not change with temperature Because of this it is sometimes more use-ful to describe solutions in terms of how many moles of solute are dissolvedin a specific mass of solvent Such a description is called molalitymdashthe ratioof the number of moles of solute dissolved in one kilogram of solvent Theunit m is read as molal A solution containing one mole of solute per kilo-gram of solvent is a one molal solution

Molality (m) moles of solute1000 g of solvent

moles of solutekilogram of solvent

For more practice withmolality problems go to SupplementalSupplementalPractice ProblemsPractice Problems

in Appendix A

Practice

EXAMPLE PROBLEM 15-5

Calculating MolalityIn the lab a student adds 45 g of sodium chloride (NaCl) to 1000 g ofwater Calculate the molality of the solution

1 Analyze the ProblemYou are given the mass of solute and solvent The molar mass of thesolute can be used to determine the number of moles of solute insolution Then the molality can be calculated

Known Unknown

mass of water (H2O) 1000 g m molkg mass of sodium chloride (NaCl) 45 g

2 Solve for the UnknownUse molar mass to calculate the number of moles of NaCl

45 g NaCl 0077 mol NaCl

Convert the mass of H2O from grams to kilograms

1000 g H2O 1100

k0ggHH2O

2O 01000 kg H2O

Substitute the known values into the expression for molality and solve

m 000170700

mkogl N

Ha2OCl

m 077 molkg

3 Evaluate the AnswerBecause there was less than one-tenth mole of solute present in one-tenth kilogram of water the molality should be less than one as it isThe mass of sodium chloride was given with two significant figurestherefore the molality also is expressed with two significant figures

moles of solutekilogram of solvent

PRACTICE PROBLEMS24 What is the molality of a solution containing 100 g Na2SO4 dissolved

in 10000 g of water

25 What is the molality of a solution containing 300 g of naphthalene(C10H8) dissolved in 5000 g of toluene

The concentration of the result-ing solution can be expressed interms of moles of solute per vol-ume (molarity) or moles of soluteper mass (molality)

1 mol NaCl5844 g NaCl

Chem MC-469

null

5135697

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

470 Chapter 15 Solutions

Section 152 Assessment

28 Distinguish between a dilute solution and a con-centrated solution

29 Compare and contrast five quantitative ways todescribe the composition of solutions

30 Describe the laboratory procedure for preparing aspecific volume of a dilute solution from a con-centrated stock solution

31 Thinking Critically Explain the similarities anddifferences between a 1M solution of NaOH and a1m solution of NaOH

32 Using Numbers A can of chicken broth contains450 mg of sodium chloride in 2400 g of brothWhat is the percent by mass of sodium chloride inthe broth

Figure 15-15

The mole fraction expresses thenumber of moles of solvent andsolute relative to the total num-ber of moles of solution Eachmole fraction can be thought ofas a percent For example themole fraction of water (XH2O) is0771 which is equivalent tosaying the solution contains771 water (on a mole basis)

Hydrochloric Acidin Aqueous Solution

XHCI XH20 10000229 0771 1000

625H20

375HCI

Mole fraction If you know the number of moles of solute and solvent youcan also express the concentration of a solution in what is known as a molefractionmdashthe ratio of the number of moles of solute in solution to the totalnumber of moles of solute and solvent

The symbol X is commonly used for mole fraction with a subscript to indi-cate the solvent or solute The mole fraction for the solvent (XA) and the molefraction for the solute (XB) can be expressed as follows

XA nA

n

A

nB XB nA

n

B

nB

where nA is the number of moles of solvent and nB is the number of molesof solute Why must the sum of the mole fractions for all components in asolution equal one

Consider as an example the mole fraction of hydrochloric acid (HCl) inthe aqueous solution shown in Figure 15-15 For every 100 grams of solu-tion 375 g would be HCl and 625 g would be H2O To convert these massesto moles you would use the molar masses as conversion factors

nHCl 375 g HCl 316m5

ogl H

HCC

ll 103 mol HCl

nH2O 625 g H2O 118m0

ogl H

H2

2

OO 347 mol H2O

Thus the mole fractions of hydrochloric acid and water can be expressed as

XHCl nHCl

nHC

nl

H2O

XHCl 0229

XH2O nHCl

n

H2O

nH2O

XH2O 0771

347 mol H2O103 mol HCl 347 mol H2O

103 mol HCl103 mol HCl 347 mol H2O

PRACTICE PROBLEMS26 What is the mole fraction of NaOH in an aqueous solution that con-

tains 228 NaOH by mass

27 An aqueous solution of NaCl has a mole fraction of 021 What is themass of NaCl dissolved in 1000 mL of solution

For more practicewith mole fractionproblems go toSupplemental

Problems in Appendix A

Practice

chemistrymccomself_check_quiz

153 Colligative Properties of Solutions 471

Objectivesbull Explain the nature of col-

ligative properties

bull Describe four colligativeproperties of solutions

bull Calculate the boiling pointelevation and the freezingpoint depression of a solution

Vocabularycolligative propertyvapor pressure loweringboiling point elevationfreezing point depressionosmosisosmotic pressure

Section 153

Colligative Properties ofSolutions

Solutes affect some of the physical properties of their solvents Earlyresearchers were puzzled to discover that the effects of a solute on a solventdepended only on how many solute particles were in solution not on the spe-cific solute dissolved Physical properties of solutions that are affected by thenumber of particles but not the identity of dissolved solute particles are calledcolligative properties The word colligative means ldquodepending on the col-lectionrdquo Colligative properties include vapor pressure lowering boiling pointelevation freezing point depression and osmotic pressure

Electrolytes and Colligative Properties In Chapter 8 you learned that ionic compounds are called electrolytes becausethey dissociate in water to form a solution that conducts electric currentSome molecular compounds ionize in water and also are electrolytesElectrolytes that produce many ions in solution are called strong electrolytesthose that produce only a few ions in solution are called weak electrolytes

Sodium chloride is a strong electrolyte It almost completely dissociatesin solution producing Na and Cl ions

NaCl(s) rarr Na(aq) Cl(aq)

Dissolving one mole of NaCl in a kilogram of water does not yield a 1m solu-tion of ions Rather there would be almost two moles of solute particles insolutionmdashapproximately one mole each of Na and Cl ions How manymoles of ions would you expect to find in a 1m aqueous solution of HCl

Nonelectrolytes in aqueous solution Many molecular compounds dis-solve in solvents but do not ionize Such solutions do not conduct an electriccurrent and the solutes are called nonelectrolytes Sucrose is an example ofa nonelectrolyte A 1m sucrose solution contains only one mole of sucroseparticles Figure 15-16 compares the conductivity of a solution containingan electrolyte solute with one containing a nonelectrolyte solute Which com-pound would have the greater effect on colligative properties sodium chlo-ride or sucrose

Figure 15-16

Sodium chloride is a strongelectrolyte and conducts electricity well Sucrosewhile soluble in water does notionize and therefore does notconduct electricity Which solutesodium chloride or sucrose pro-duces more particles in solutionper mole

b

a

Cl

Na

H2O

C12H22O11

H2O

a b

Chem MC-471

null

15662985

CMC_Spn_C15-S03

null

335933

Vapor Pressure LoweringIn Chapter 13 you learned that vapor pressure is the pressure exerted in aclosed container by liquid particles that have escaped the liquidrsquos surface andentered the gaseous state In a closed container at constant temperature andpressure the solvent particles reach a state of dynamic equilibrium escapingand reentering the liquid state at the same rate The vapor pressure for a closedcontainer of pure water is shown in Figure 15-17a

Experiments show that adding a nonvolatile solute (one that has little ten-dency to become a gas) to a solvent lowers the solventrsquos vapor pressure Theparticles that produce vapor pressure escape the liquid phase at its surfaceWhen a solvent is pure as shown in Figure 15-17a its particles occupy theentire surface area However when the solvent contains solute as shown inFigure 15-17b a mix of solute and solvent particles occupies the surfacearea With fewer solvent particles at the surface fewer particles enter thegaseous state and the vapor pressure is lowered The greater the number ofsolute particles in a solvent the lower the resulting vapor pressure Thusvapor pressure lowering is due to the number of solute particles in solutionand is a colligative property of solutions

You can predict the relative effect of a solute on vapor pressure based onwhether the solute is an electrolyte or a nonelectrolyte For example one moleeach of the solvated nonelectrolytes glucose sucrose and ethanol moleculeshas the same relative effect on the vapor pressure However one mole eachof the solvated electrolytes sodium chloride sodium sulfate and aluminumchloride has an increasingly greater effect on vapor pressure because of theincreasing number of ions each produces in solution

Boiling Point ElevationBecause a solute lowers a solventrsquos vapor pressure it also affects the boilingpoint of the solvent Recall from Chapter 13 that the liquid in a pot on yourstove boils when its vapor pressure equals atmospheric pressure When thetemperature of a solution containing a nonvolatile solute is raised to the boil-ing point of the pure solvent the resulting vapor pressure is still less than theatmospheric pressure and the solution will not boil Thus the solution mustbe heated to a higher temperature to supply the additional kinetic energyneeded to raise the vapor pressure to atmospheric pressure The temperaturedifference between a solutionrsquos boiling point and a pure solventrsquos boiling pointis called the boiling point elevation

For nonelectrolytes the value of the boiling point elevation which is sym-bolized Tb is directly proportional to the solutionrsquos molality

Tb = Kbm

The molal boiling point elevation constant Kb is the difference in boilingpoints between a 1m nonvolatile nonelectrolyte solution and a pure solventIt is expressed in units of degCm and varies for different solvents Values of Kbfor several common solvents are found in Table 15-4 Note that waterrsquos Kbvalue is 0512degCm This means that a 1m aqueous solution containing a non-volatile nonelectrolyte solute boils at 100512degC a temperature 0512degChigher than pure waterrsquos boiling point of 1000degC

Like vapor pressure lowering boiling point elevation is a colligative prop-erty The value of the boiling point elevation is directly proportional to thesolutionrsquos solute molality that is the greater the number of solute particlesin the solution the greater the boiling point elevation

472 Chapter 15 Solutions

Figure 15-17

The vapor pressure of a pure solvent is greater than thevapor pressure of a solution con-taining a nonvolatile solute b

a

Water

a

b

Molal Boiling-PointElevation Constants (Kb)

Boiling KbSolvent point (degC) (degCm)

Water 1000 0512

Benzene 801 253

Carbontetrachloride 767 503

Ethanol 785 122

Chloroform 617 363

Table 15-4

Sucrose

Chem MC-472

null

23427336

Freezing Point DepressionThe freezing point depression of a solution is anothercolligative property of solutions At a solventrsquos freez-ing point temperature the particles no longer havesufficient kinetic energy to overcome the interparti-cle attractive forces the particles form into a moreorganized structure in the solid state In a solutionthe solute particles interfere with the attractive forcesamong the solvent particles This prevents the sol-vent from entering the solid state at its normal freez-ing point The freezing point of a solution is alwayslower than that of a pure solvent Figure 15-18shows the differences in boiling and melting pointsof pure water and an aqueous solution By compar-ing the solid and dashed lines you can see that thetemperature range over which the aqueous solutionexists as a liquid is greater than that of pure waterYou can observe the effect of freezing point depres-sion by doing the miniLAB on this page

A solutionrsquos freezing point depression Tf is the differencein temperature between its freezing point and the freezing pointof its pure solvent For nonelectrolytes the value of the freezingpoint depression which is symbolized as Tf is directly propor-tional to the solutionrsquos molality

Tf = Kfm

153 Colligative Properties of Solutions 473

Pure solventSolution

LIQUIDSOLID

GAS

P

Tf Tb

Increasing Temperature

Freezing pointof solution

1 atm

Boiling pointof solution

Normal freezing point of water

Normal boilingpoint of water

Incr

easi

ng

Pre

ssu

re

Freezing Point DepressionMeasuring The colligative property of freezingpoint depression can be observed in a simple lab-oratory investigation You will measure the tem-peratures of two beakers and their contents

Materials 400-mL beakers (2) crushed ice rocksalt (NaCl) water stirring rods (2) graduatedcylinder thermometers (2) balance

Procedure1 Fill two 400-mL beakers with crushed ice Add

50 mL of cold tap water to each beaker

2 Stir the contents of each beaker with a stirringrod until both beakers are at a constant tem-perature approximately one minute

3 Measure the temperature of each beaker usinga thermometer and record the readings

4 Add 75 g of rock salt to one of the beakersContinue stirring both beakers Some of thesalt will dissolve

5 When the temperature in each beaker is con-stant record the readings

6 To clean up flush the contents of each beakerdown the drain with excess water

Analysis1 Compare your readings taken for the ice water

and the salt water How do you explain theobserved temperature change

2 Why was salt only added to one of the beakers

3 Salt is a strong electrolyte that produces twoions Na and Cl when it dissociates in waterWhy is this important to consider when calcu-lating the colligative property of freezingpoint depression

4 Predict if it would it be better to use coarserock salt or fine table salt when making home-made ice cream Explain

miniLAB

Figure 15-18

This phase diagram shows how tempera-ture and pressure affect the solid liquidand gas phases of a pure solvent (solidlines) and a solution (dashed lines) Thedifference between the solid and dashedlines corresponds to vapor pressure lower-ing (P) boiling point elevation (Tb ) and freezing point depression (Tf)

Chem MC-473

null

836437

474 Chapter 15 Solutions

EXAMPLE PROBLEM 15-6

Changes in Boiling and Freezing PointsWhat are the boiling point and freezing point of a 0029m aqueous solu-tion of sodium chloride (NaCl)

1 Analyze the ProblemYou are given the molality of an aqueous sodium chloride solutionFirst calculate Tb and Tf based on the number of particles in solu-tion Then to determine the elevated boiling point and thedepressed freezing point add Tb to the normal boiling point andsubtract Tf from the normal freezing point

Known

solute sodium chloride (NaCl)molality of sodium chloride solution 0029m

Unknown

boiling point degCfreezing point degC

2 Solve for the UnknownEach mole of the electrolyte sodium chloride dissociates in solution toproduce two moles of particles Calculate the effective number ofsolute particles in solutionparticle molality 2 0029m 0058m

Substitute the known values for Kb Kf and particle molality into theTb and Tf equations and solve

Tb Kbm (0512degCm)(0058m) 0030degC

Tf Kfm (186degCm)(0058m) 011degC

Add Tb to the normal boiling point and subtract Tf from the nor-mal freezing point to determine the elevated boiling point anddepressed freezing point of the solution

boiling point 1000degC 0030degC 100030degC

freezing point 00degC 011degC 011degC

3 Evaluate the AnswerThe boiling point is higher and the freezing point is lower asexpected Because the molality of the solution has two significant fig-ures both Tb and Tf also have two significant figures Because thenormal boiling point and freezing point are exact values they do notaffect the number of significant figures in the final answer

Molal Freezing PointDepression Constants (Kf)

Freezing KfSolvent point (degC) (degCm)

Water 00 186

Benzene 55 512

Carbontetrachloride ndash23 298

Ethanol ndash1141 199

Chloroform ndash635 468

Table 15-5

Values of Kf for several common solvents are found in Table 15-5 As withKb values Kf values are specific to their solvents With waterrsquos Kf value of186degCm a 1m aqueous solution containing a nonvolatile nonelectrolytesolute freezes at ndash186degC rather than at pure waterrsquos freezing point of 00degC

However in using the Tb and Tf relationships with electrolytes youmust make sure to use the effective molality of the solution For example onemole of the electrolyte sodium chloride (NaCl) forms one mole of Na+ ionsand one mole of Clndash ions in solution Thus a 1m aqueous NaCl solution pro-duces two moles of solute particles in a kilogram of watermdasheffectively act-ing as a 2m solution Because of this 2m and not 1m must be used in theTb and Tf relationships for a 1m solution of sodium chloride The fol-lowing Example Problem illustrates this point

Chem MC-474

null

7491926

153 Colligative Properties of Solutions 475

Section 153 Assessment

37 Explain the nature of colligative properties

38 Describe four colligative properties of solutions

39 Explain why a solution has a lower boiling pointthan the pure solvent

40 Thinking Critically Explain why the colligativeproperties described in this section may not apply

to solutions containing volatile solutes HintVolatile solutes are able to leave the liquid phaseand enter the gas phase

41 Using Numbers Calculate the boiling point ele-vation and freezing point depression of a solutioncontaining 500 g of glucose (C6H12O6) dissolvedin 5000 g of water

Osmosis and Osmotic PressureIn Chapter 13 you learned that diffusion is the mixing of gases or liquidsresulting from their random motions Osmosis is the diffusion of solventparticles across a semipermeable membrane from an area of higher solventconcentration to an area of lower solvent concentration Semipermeable mem-branes are barriers with tiny pores that allow some but not all kinds of parti-cles to cross The membranes surrounding all living cells are semipermeablemembranes Osmosis plays an important role in many biological systems suchas kidney dialysis and the uptake of nutrients by plants

Letrsquos look at a simple system in which a sucrose-water solution is sepa-rated from its solventmdashpure watermdashby a semipermeable membrane Duringosmosis water molecules move in both directions across the membrane butthe sugar molecules cannot cross it Both water and sugar molecules contactthe membrane on the solution side but only water molecules contact the mem-brane on the pure solvent side Thus more water molecules cross the mem-brane from the pure solvent side than from the solution side

The additional water molecules on the solution side of the membrane cre-ate pressure and push some water molecules back across the membrane Theamount of additional pressure caused by the water molecules that moved intothe solution is called the osmotic pressure Osmotic pressure depends uponthe number of solute particles in a given volume of solution Thereforeosmotic pressure is another colligative property of solutions

PRACTICE PROBLEMS33 What are the boiling point and freezing point of a 0625m aqueous

solution of any nonvolatile nonelectrolyte solute

34 What are the boiling point and freezing point of a 040m solution ofsucrose in ethanol

35 A lab technician determines the boiling point elevation of an aque-ous solution of a nonvolatile nonelectrolyte to be 112degC What is thesolutionrsquos molality

36 A student dissolves 0500 mol of a nonvolatile nonelectrolyte solute inone kilogram of benzene (C6H6) What is the boiling point elevationof the resulting solution

For more practice withboiling point elevationand freezing pointdepression problems

go to SupplementalPractice Problems inAppendix A

Practice

Renal DialysisTechnicianPeople with poorly functioningkidneys must routinely undergodialysis in order to survive Theprocess which removes impuri-ties from the blood throughosmosis can be stressful for thepatient If you enjoy workingwith and helping people andwant to work in a medical set-ting consider being a renaldialysis technician

Dialysis technicians preparethe dialysis equipment and thepatient for treatment start thetreatment and monitor theprocess They also make neces-sary adjustments keep recordsand respond to emergenciesTechnicians spend more timewith the patients than thedoctors or nurses They workin hospitals outpatient facili-ties and home-based dialysisprograms

chemistrymccomself_check_quiz

Vapor Pressure LoweringIn Chapter 13 you learned that vapor pressure is the pressure exerted in aclosed container by liquid particles that have escaped the liquidrsquos surface andentered the gaseous state In a closed container at constant temperature andpressure the solvent particles reach a state of dynamic equilibrium escapingand reentering the liquid state at the same rate The vapor pressure for a closedcontainer of pure water is shown in Figure 15-17a

Experiments show that adding a nonvolatile solute (one that has little ten-dency to become a gas) to a solvent lowers the solventrsquos vapor pressure Theparticles that produce vapor pressure escape the liquid phase at its surfaceWhen a solvent is pure as shown in Figure 15-17a its particles occupy theentire surface area However when the solvent contains solute as shown inFigure 15-17b a mix of solute and solvent particles occupies the surfacearea With fewer solvent particles at the surface fewer particles enter thegaseous state and the vapor pressure is lowered The greater the number ofsolute particles in a solvent the lower the resulting vapor pressure Thusvapor pressure lowering is due to the number of solute particles in solutionand is a colligative property of solutions

You can predict the relative effect of a solute on vapor pressure based onwhether the solute is an electrolyte or a nonelectrolyte For example one moleeach of the solvated nonelectrolytes glucose sucrose and ethanol moleculeshas the same relative effect on the vapor pressure However one mole eachof the solvated electrolytes sodium chloride sodium sulfate and aluminumchloride has an increasingly greater effect on vapor pressure because of theincreasing number of ions each produces in solution

Boiling Point ElevationBecause a solute lowers a solventrsquos vapor pressure it also affects the boilingpoint of the solvent Recall from Chapter 13 that the liquid in a pot on yourstove boils when its vapor pressure equals atmospheric pressure When thetemperature of a solution containing a nonvolatile solute is raised to the boil-ing point of the pure solvent the resulting vapor pressure is still less than theatmospheric pressure and the solution will not boil Thus the solution mustbe heated to a higher temperature to supply the additional kinetic energyneeded to raise the vapor pressure to atmospheric pressure The temperaturedifference between a solutionrsquos boiling point and a pure solventrsquos boiling pointis called the boiling point elevation

For nonelectrolytes the value of the boiling point elevation which is sym-bolized Tb is directly proportional to the solutionrsquos molality

Tb = Kbm

The molal boiling point elevation constant Kb is the difference in boilingpoints between a 1m nonvolatile nonelectrolyte solution and a pure solventIt is expressed in units of degCm and varies for different solvents Values of Kbfor several common solvents are found in Table 15-4 Note that waterrsquos Kbvalue is 0512degCm This means that a 1m aqueous solution containing a non-volatile nonelectrolyte solute boils at 100512degC a temperature 0512degChigher than pure waterrsquos boiling point of 1000degC

Like vapor pressure lowering boiling point elevation is a colligative prop-erty The value of the boiling point elevation is directly proportional to thesolutionrsquos solute molality that is the greater the number of solute particlesin the solution the greater the boiling point elevation

472 Chapter 15 Solutions

Figure 15-17

The vapor pressure of a pure solvent is greater than thevapor pressure of a solution con-taining a nonvolatile solute b

a

Water

a

b

Molal Boiling-PointElevation Constants (Kb)

Boiling KbSolvent point (degC) (degCm)

Water 1000 0512

Benzene 801 253

Carbontetrachloride 767 503

Ethanol 785 122

Chloroform 617 363

Table 15-4

Sucrose

Chem MC-472

null

23427336

Freezing Point DepressionThe freezing point depression of a solution is anothercolligative property of solutions At a solventrsquos freez-ing point temperature the particles no longer havesufficient kinetic energy to overcome the interparti-cle attractive forces the particles form into a moreorganized structure in the solid state In a solutionthe solute particles interfere with the attractive forcesamong the solvent particles This prevents the sol-vent from entering the solid state at its normal freez-ing point The freezing point of a solution is alwayslower than that of a pure solvent Figure 15-18shows the differences in boiling and melting pointsof pure water and an aqueous solution By compar-ing the solid and dashed lines you can see that thetemperature range over which the aqueous solutionexists as a liquid is greater than that of pure waterYou can observe the effect of freezing point depres-sion by doing the miniLAB on this page

A solutionrsquos freezing point depression Tf is the differencein temperature between its freezing point and the freezing pointof its pure solvent For nonelectrolytes the value of the freezingpoint depression which is symbolized as Tf is directly propor-tional to the solutionrsquos molality

Tf = Kfm

153 Colligative Properties of Solutions 473

Pure solventSolution

LIQUIDSOLID

GAS

P

Tf Tb

Increasing Temperature

Freezing pointof solution

1 atm

Boiling pointof solution

Normal freezing point of water

Normal boilingpoint of water

Incr

easi

ng

Pre

ssu

re

Freezing Point DepressionMeasuring The colligative property of freezingpoint depression can be observed in a simple lab-oratory investigation You will measure the tem-peratures of two beakers and their contents

Materials 400-mL beakers (2) crushed ice rocksalt (NaCl) water stirring rods (2) graduatedcylinder thermometers (2) balance

Procedure1 Fill two 400-mL beakers with crushed ice Add

50 mL of cold tap water to each beaker

2 Stir the contents of each beaker with a stirringrod until both beakers are at a constant tem-perature approximately one minute

3 Measure the temperature of each beaker usinga thermometer and record the readings

4 Add 75 g of rock salt to one of the beakersContinue stirring both beakers Some of thesalt will dissolve

5 When the temperature in each beaker is con-stant record the readings

6 To clean up flush the contents of each beakerdown the drain with excess water

Analysis1 Compare your readings taken for the ice water

and the salt water How do you explain theobserved temperature change

2 Why was salt only added to one of the beakers

3 Salt is a strong electrolyte that produces twoions Na and Cl when it dissociates in waterWhy is this important to consider when calcu-lating the colligative property of freezingpoint depression

4 Predict if it would it be better to use coarserock salt or fine table salt when making home-made ice cream Explain

miniLAB

Figure 15-18

This phase diagram shows how tempera-ture and pressure affect the solid liquidand gas phases of a pure solvent (solidlines) and a solution (dashed lines) Thedifference between the solid and dashedlines corresponds to vapor pressure lower-ing (P) boiling point elevation (Tb ) and freezing point depression (Tf)

Chem MC-473

null

836437

474 Chapter 15 Solutions

EXAMPLE PROBLEM 15-6

Changes in Boiling and Freezing PointsWhat are the boiling point and freezing point of a 0029m aqueous solu-tion of sodium chloride (NaCl)

1 Analyze the ProblemYou are given the molality of an aqueous sodium chloride solutionFirst calculate Tb and Tf based on the number of particles in solu-tion Then to determine the elevated boiling point and thedepressed freezing point add Tb to the normal boiling point andsubtract Tf from the normal freezing point

Known

solute sodium chloride (NaCl)molality of sodium chloride solution 0029m

Unknown

boiling point degCfreezing point degC

2 Solve for the UnknownEach mole of the electrolyte sodium chloride dissociates in solution toproduce two moles of particles Calculate the effective number ofsolute particles in solutionparticle molality 2 0029m 0058m

Substitute the known values for Kb Kf and particle molality into theTb and Tf equations and solve

Tb Kbm (0512degCm)(0058m) 0030degC

Tf Kfm (186degCm)(0058m) 011degC

Add Tb to the normal boiling point and subtract Tf from the nor-mal freezing point to determine the elevated boiling point anddepressed freezing point of the solution

boiling point 1000degC 0030degC 100030degC

freezing point 00degC 011degC 011degC

3 Evaluate the AnswerThe boiling point is higher and the freezing point is lower asexpected Because the molality of the solution has two significant fig-ures both Tb and Tf also have two significant figures Because thenormal boiling point and freezing point are exact values they do notaffect the number of significant figures in the final answer

Molal Freezing PointDepression Constants (Kf)

Freezing KfSolvent point (degC) (degCm)

Water 00 186

Benzene 55 512

Carbontetrachloride ndash23 298

Ethanol ndash1141 199

Chloroform ndash635 468

Table 15-5

Values of Kf for several common solvents are found in Table 15-5 As withKb values Kf values are specific to their solvents With waterrsquos Kf value of186degCm a 1m aqueous solution containing a nonvolatile nonelectrolytesolute freezes at ndash186degC rather than at pure waterrsquos freezing point of 00degC

However in using the Tb and Tf relationships with electrolytes youmust make sure to use the effective molality of the solution For example onemole of the electrolyte sodium chloride (NaCl) forms one mole of Na+ ionsand one mole of Clndash ions in solution Thus a 1m aqueous NaCl solution pro-duces two moles of solute particles in a kilogram of watermdasheffectively act-ing as a 2m solution Because of this 2m and not 1m must be used in theTb and Tf relationships for a 1m solution of sodium chloride The fol-lowing Example Problem illustrates this point

Chem MC-474

null

7491926

153 Colligative Properties of Solutions 475

Section 153 Assessment

37 Explain the nature of colligative properties

38 Describe four colligative properties of solutions

39 Explain why a solution has a lower boiling pointthan the pure solvent

40 Thinking Critically Explain why the colligativeproperties described in this section may not apply

to solutions containing volatile solutes HintVolatile solutes are able to leave the liquid phaseand enter the gas phase

41 Using Numbers Calculate the boiling point ele-vation and freezing point depression of a solutioncontaining 500 g of glucose (C6H12O6) dissolvedin 5000 g of water

Osmosis and Osmotic PressureIn Chapter 13 you learned that diffusion is the mixing of gases or liquidsresulting from their random motions Osmosis is the diffusion of solventparticles across a semipermeable membrane from an area of higher solventconcentration to an area of lower solvent concentration Semipermeable mem-branes are barriers with tiny pores that allow some but not all kinds of parti-cles to cross The membranes surrounding all living cells are semipermeablemembranes Osmosis plays an important role in many biological systems suchas kidney dialysis and the uptake of nutrients by plants

Letrsquos look at a simple system in which a sucrose-water solution is sepa-rated from its solventmdashpure watermdashby a semipermeable membrane Duringosmosis water molecules move in both directions across the membrane butthe sugar molecules cannot cross it Both water and sugar molecules contactthe membrane on the solution side but only water molecules contact the mem-brane on the pure solvent side Thus more water molecules cross the mem-brane from the pure solvent side than from the solution side

The additional water molecules on the solution side of the membrane cre-ate pressure and push some water molecules back across the membrane Theamount of additional pressure caused by the water molecules that moved intothe solution is called the osmotic pressure Osmotic pressure depends uponthe number of solute particles in a given volume of solution Thereforeosmotic pressure is another colligative property of solutions

PRACTICE PROBLEMS33 What are the boiling point and freezing point of a 0625m aqueous

solution of any nonvolatile nonelectrolyte solute

34 What are the boiling point and freezing point of a 040m solution ofsucrose in ethanol

35 A lab technician determines the boiling point elevation of an aque-ous solution of a nonvolatile nonelectrolyte to be 112degC What is thesolutionrsquos molality

36 A student dissolves 0500 mol of a nonvolatile nonelectrolyte solute inone kilogram of benzene (C6H6) What is the boiling point elevationof the resulting solution

For more practice withboiling point elevationand freezing pointdepression problems

go to SupplementalPractice Problems inAppendix A

Practice

Renal DialysisTechnicianPeople with poorly functioningkidneys must routinely undergodialysis in order to survive Theprocess which removes impuri-ties from the blood throughosmosis can be stressful for thepatient If you enjoy workingwith and helping people andwant to work in a medical set-ting consider being a renaldialysis technician

Dialysis technicians preparethe dialysis equipment and thepatient for treatment start thetreatment and monitor theprocess They also make neces-sary adjustments keep recordsand respond to emergenciesTechnicians spend more timewith the patients than thedoctors or nurses They workin hospitals outpatient facili-ties and home-based dialysisprograms

chemistrymccomself_check_quiz

Freezing Point DepressionThe freezing point depression of a solution is anothercolligative property of solutions At a solventrsquos freez-ing point temperature the particles no longer havesufficient kinetic energy to overcome the interparti-cle attractive forces the particles form into a moreorganized structure in the solid state In a solutionthe solute particles interfere with the attractive forcesamong the solvent particles This prevents the sol-vent from entering the solid state at its normal freez-ing point The freezing point of a solution is alwayslower than that of a pure solvent Figure 15-18shows the differences in boiling and melting pointsof pure water and an aqueous solution By compar-ing the solid and dashed lines you can see that thetemperature range over which the aqueous solutionexists as a liquid is greater than that of pure waterYou can observe the effect of freezing point depres-sion by doing the miniLAB on this page

A solutionrsquos freezing point depression Tf is the differencein temperature between its freezing point and the freezing pointof its pure solvent For nonelectrolytes the value of the freezingpoint depression which is symbolized as Tf is directly propor-tional to the solutionrsquos molality

Tf = Kfm

153 Colligative Properties of Solutions 473

Pure solventSolution

LIQUIDSOLID

GAS

P

Tf Tb

Increasing Temperature

Freezing pointof solution

1 atm

Boiling pointof solution

Normal freezing point of water

Normal boilingpoint of water

Incr

easi

ng

Pre

ssu

re

Freezing Point DepressionMeasuring The colligative property of freezingpoint depression can be observed in a simple lab-oratory investigation You will measure the tem-peratures of two beakers and their contents

Materials 400-mL beakers (2) crushed ice rocksalt (NaCl) water stirring rods (2) graduatedcylinder thermometers (2) balance

Procedure1 Fill two 400-mL beakers with crushed ice Add

50 mL of cold tap water to each beaker

2 Stir the contents of each beaker with a stirringrod until both beakers are at a constant tem-perature approximately one minute

3 Measure the temperature of each beaker usinga thermometer and record the readings

4 Add 75 g of rock salt to one of the beakersContinue stirring both beakers Some of thesalt will dissolve

5 When the temperature in each beaker is con-stant record the readings

6 To clean up flush the contents of each beakerdown the drain with excess water

Analysis1 Compare your readings taken for the ice water

and the salt water How do you explain theobserved temperature change

2 Why was salt only added to one of the beakers

3 Salt is a strong electrolyte that produces twoions Na and Cl when it dissociates in waterWhy is this important to consider when calcu-lating the colligative property of freezingpoint depression

4 Predict if it would it be better to use coarserock salt or fine table salt when making home-made ice cream Explain

miniLAB

Figure 15-18

This phase diagram shows how tempera-ture and pressure affect the solid liquidand gas phases of a pure solvent (solidlines) and a solution (dashed lines) Thedifference between the solid and dashedlines corresponds to vapor pressure lower-ing (P) boiling point elevation (Tb ) and freezing point depression (Tf)

Chem MC-473

null

836437

474 Chapter 15 Solutions

EXAMPLE PROBLEM 15-6

Changes in Boiling and Freezing PointsWhat are the boiling point and freezing point of a 0029m aqueous solu-tion of sodium chloride (NaCl)

1 Analyze the ProblemYou are given the molality of an aqueous sodium chloride solutionFirst calculate Tb and Tf based on the number of particles in solu-tion Then to determine the elevated boiling point and thedepressed freezing point add Tb to the normal boiling point andsubtract Tf from the normal freezing point

Known

solute sodium chloride (NaCl)molality of sodium chloride solution 0029m

Unknown

boiling point degCfreezing point degC

2 Solve for the UnknownEach mole of the electrolyte sodium chloride dissociates in solution toproduce two moles of particles Calculate the effective number ofsolute particles in solutionparticle molality 2 0029m 0058m

Substitute the known values for Kb Kf and particle molality into theTb and Tf equations and solve

Tb Kbm (0512degCm)(0058m) 0030degC

Tf Kfm (186degCm)(0058m) 011degC

Add Tb to the normal boiling point and subtract Tf from the nor-mal freezing point to determine the elevated boiling point anddepressed freezing point of the solution

boiling point 1000degC 0030degC 100030degC

freezing point 00degC 011degC 011degC

3 Evaluate the AnswerThe boiling point is higher and the freezing point is lower asexpected Because the molality of the solution has two significant fig-ures both Tb and Tf also have two significant figures Because thenormal boiling point and freezing point are exact values they do notaffect the number of significant figures in the final answer

Molal Freezing PointDepression Constants (Kf)

Freezing KfSolvent point (degC) (degCm)

Water 00 186

Benzene 55 512

Carbontetrachloride ndash23 298

Ethanol ndash1141 199

Chloroform ndash635 468

Table 15-5

Values of Kf for several common solvents are found in Table 15-5 As withKb values Kf values are specific to their solvents With waterrsquos Kf value of186degCm a 1m aqueous solution containing a nonvolatile nonelectrolytesolute freezes at ndash186degC rather than at pure waterrsquos freezing point of 00degC

However in using the Tb and Tf relationships with electrolytes youmust make sure to use the effective molality of the solution For example onemole of the electrolyte sodium chloride (NaCl) forms one mole of Na+ ionsand one mole of Clndash ions in solution Thus a 1m aqueous NaCl solution pro-duces two moles of solute particles in a kilogram of watermdasheffectively act-ing as a 2m solution Because of this 2m and not 1m must be used in theTb and Tf relationships for a 1m solution of sodium chloride The fol-lowing Example Problem illustrates this point

Chem MC-474

null

7491926

153 Colligative Properties of Solutions 475

Section 153 Assessment

37 Explain the nature of colligative properties

38 Describe four colligative properties of solutions

39 Explain why a solution has a lower boiling pointthan the pure solvent

40 Thinking Critically Explain why the colligativeproperties described in this section may not apply

to solutions containing volatile solutes HintVolatile solutes are able to leave the liquid phaseand enter the gas phase

41 Using Numbers Calculate the boiling point ele-vation and freezing point depression of a solutioncontaining 500 g of glucose (C6H12O6) dissolvedin 5000 g of water

Osmosis and Osmotic PressureIn Chapter 13 you learned that diffusion is the mixing of gases or liquidsresulting from their random motions Osmosis is the diffusion of solventparticles across a semipermeable membrane from an area of higher solventconcentration to an area of lower solvent concentration Semipermeable mem-branes are barriers with tiny pores that allow some but not all kinds of parti-cles to cross The membranes surrounding all living cells are semipermeablemembranes Osmosis plays an important role in many biological systems suchas kidney dialysis and the uptake of nutrients by plants

Letrsquos look at a simple system in which a sucrose-water solution is sepa-rated from its solventmdashpure watermdashby a semipermeable membrane Duringosmosis water molecules move in both directions across the membrane butthe sugar molecules cannot cross it Both water and sugar molecules contactthe membrane on the solution side but only water molecules contact the mem-brane on the pure solvent side Thus more water molecules cross the mem-brane from the pure solvent side than from the solution side

The additional water molecules on the solution side of the membrane cre-ate pressure and push some water molecules back across the membrane Theamount of additional pressure caused by the water molecules that moved intothe solution is called the osmotic pressure Osmotic pressure depends uponthe number of solute particles in a given volume of solution Thereforeosmotic pressure is another colligative property of solutions

PRACTICE PROBLEMS33 What are the boiling point and freezing point of a 0625m aqueous

solution of any nonvolatile nonelectrolyte solute

34 What are the boiling point and freezing point of a 040m solution ofsucrose in ethanol

35 A lab technician determines the boiling point elevation of an aque-ous solution of a nonvolatile nonelectrolyte to be 112degC What is thesolutionrsquos molality

36 A student dissolves 0500 mol of a nonvolatile nonelectrolyte solute inone kilogram of benzene (C6H6) What is the boiling point elevationof the resulting solution

For more practice withboiling point elevationand freezing pointdepression problems

go to SupplementalPractice Problems inAppendix A

Practice

Renal DialysisTechnicianPeople with poorly functioningkidneys must routinely undergodialysis in order to survive Theprocess which removes impuri-ties from the blood throughosmosis can be stressful for thepatient If you enjoy workingwith and helping people andwant to work in a medical set-ting consider being a renaldialysis technician

Dialysis technicians preparethe dialysis equipment and thepatient for treatment start thetreatment and monitor theprocess They also make neces-sary adjustments keep recordsand respond to emergenciesTechnicians spend more timewith the patients than thedoctors or nurses They workin hospitals outpatient facili-ties and home-based dialysisprograms

chemistrymccomself_check_quiz

474 Chapter 15 Solutions

EXAMPLE PROBLEM 15-6

Changes in Boiling and Freezing PointsWhat are the boiling point and freezing point of a 0029m aqueous solu-tion of sodium chloride (NaCl)

1 Analyze the ProblemYou are given the molality of an aqueous sodium chloride solutionFirst calculate Tb and Tf based on the number of particles in solu-tion Then to determine the elevated boiling point and thedepressed freezing point add Tb to the normal boiling point andsubtract Tf from the normal freezing point

Known

solute sodium chloride (NaCl)molality of sodium chloride solution 0029m

Unknown

boiling point degCfreezing point degC

2 Solve for the UnknownEach mole of the electrolyte sodium chloride dissociates in solution toproduce two moles of particles Calculate the effective number ofsolute particles in solutionparticle molality 2 0029m 0058m

Substitute the known values for Kb Kf and particle molality into theTb and Tf equations and solve

Tb Kbm (0512degCm)(0058m) 0030degC

Tf Kfm (186degCm)(0058m) 011degC

Add Tb to the normal boiling point and subtract Tf from the nor-mal freezing point to determine the elevated boiling point anddepressed freezing point of the solution

boiling point 1000degC 0030degC 100030degC

freezing point 00degC 011degC 011degC

3 Evaluate the AnswerThe boiling point is higher and the freezing point is lower asexpected Because the molality of the solution has two significant fig-ures both Tb and Tf also have two significant figures Because thenormal boiling point and freezing point are exact values they do notaffect the number of significant figures in the final answer

Molal Freezing PointDepression Constants (Kf)

Freezing KfSolvent point (degC) (degCm)

Water 00 186

Benzene 55 512

Carbontetrachloride ndash23 298

Ethanol ndash1141 199

Chloroform ndash635 468

Table 15-5

Values of Kf for several common solvents are found in Table 15-5 As withKb values Kf values are specific to their solvents With waterrsquos Kf value of186degCm a 1m aqueous solution containing a nonvolatile nonelectrolytesolute freezes at ndash186degC rather than at pure waterrsquos freezing point of 00degC

However in using the Tb and Tf relationships with electrolytes youmust make sure to use the effective molality of the solution For example onemole of the electrolyte sodium chloride (NaCl) forms one mole of Na+ ionsand one mole of Clndash ions in solution Thus a 1m aqueous NaCl solution pro-duces two moles of solute particles in a kilogram of watermdasheffectively act-ing as a 2m solution Because of this 2m and not 1m must be used in theTb and Tf relationships for a 1m solution of sodium chloride The fol-lowing Example Problem illustrates this point

Chem MC-474

null

7491926

153 Colligative Properties of Solutions 475

Section 153 Assessment

37 Explain the nature of colligative properties

38 Describe four colligative properties of solutions

39 Explain why a solution has a lower boiling pointthan the pure solvent

40 Thinking Critically Explain why the colligativeproperties described in this section may not apply

to solutions containing volatile solutes HintVolatile solutes are able to leave the liquid phaseand enter the gas phase

41 Using Numbers Calculate the boiling point ele-vation and freezing point depression of a solutioncontaining 500 g of glucose (C6H12O6) dissolvedin 5000 g of water

Osmosis and Osmotic PressureIn Chapter 13 you learned that diffusion is the mixing of gases or liquidsresulting from their random motions Osmosis is the diffusion of solventparticles across a semipermeable membrane from an area of higher solventconcentration to an area of lower solvent concentration Semipermeable mem-branes are barriers with tiny pores that allow some but not all kinds of parti-cles to cross The membranes surrounding all living cells are semipermeablemembranes Osmosis plays an important role in many biological systems suchas kidney dialysis and the uptake of nutrients by plants

Letrsquos look at a simple system in which a sucrose-water solution is sepa-rated from its solventmdashpure watermdashby a semipermeable membrane Duringosmosis water molecules move in both directions across the membrane butthe sugar molecules cannot cross it Both water and sugar molecules contactthe membrane on the solution side but only water molecules contact the mem-brane on the pure solvent side Thus more water molecules cross the mem-brane from the pure solvent side than from the solution side

The additional water molecules on the solution side of the membrane cre-ate pressure and push some water molecules back across the membrane Theamount of additional pressure caused by the water molecules that moved intothe solution is called the osmotic pressure Osmotic pressure depends uponthe number of solute particles in a given volume of solution Thereforeosmotic pressure is another colligative property of solutions

PRACTICE PROBLEMS33 What are the boiling point and freezing point of a 0625m aqueous

solution of any nonvolatile nonelectrolyte solute

34 What are the boiling point and freezing point of a 040m solution ofsucrose in ethanol

35 A lab technician determines the boiling point elevation of an aque-ous solution of a nonvolatile nonelectrolyte to be 112degC What is thesolutionrsquos molality

36 A student dissolves 0500 mol of a nonvolatile nonelectrolyte solute inone kilogram of benzene (C6H6) What is the boiling point elevationof the resulting solution

For more practice withboiling point elevationand freezing pointdepression problems

go to SupplementalPractice Problems inAppendix A

Practice

Renal DialysisTechnicianPeople with poorly functioningkidneys must routinely undergodialysis in order to survive Theprocess which removes impuri-ties from the blood throughosmosis can be stressful for thepatient If you enjoy workingwith and helping people andwant to work in a medical set-ting consider being a renaldialysis technician

Dialysis technicians preparethe dialysis equipment and thepatient for treatment start thetreatment and monitor theprocess They also make neces-sary adjustments keep recordsand respond to emergenciesTechnicians spend more timewith the patients than thedoctors or nurses They workin hospitals outpatient facili-ties and home-based dialysisprograms

chemistrymccomself_check_quiz

153 Colligative Properties of Solutions 475

Section 153 Assessment

37 Explain the nature of colligative properties

38 Describe four colligative properties of solutions

39 Explain why a solution has a lower boiling pointthan the pure solvent

40 Thinking Critically Explain why the colligativeproperties described in this section may not apply

to solutions containing volatile solutes HintVolatile solutes are able to leave the liquid phaseand enter the gas phase

41 Using Numbers Calculate the boiling point ele-vation and freezing point depression of a solutioncontaining 500 g of glucose (C6H12O6) dissolvedin 5000 g of water

Osmosis and Osmotic PressureIn Chapter 13 you learned that diffusion is the mixing of gases or liquidsresulting from their random motions Osmosis is the diffusion of solventparticles across a semipermeable membrane from an area of higher solventconcentration to an area of lower solvent concentration Semipermeable mem-branes are barriers with tiny pores that allow some but not all kinds of parti-cles to cross The membranes surrounding all living cells are semipermeablemembranes Osmosis plays an important role in many biological systems suchas kidney dialysis and the uptake of nutrients by plants

Letrsquos look at a simple system in which a sucrose-water solution is sepa-rated from its solventmdashpure watermdashby a semipermeable membrane Duringosmosis water molecules move in both directions across the membrane butthe sugar molecules cannot cross it Both water and sugar molecules contactthe membrane on the solution side but only water molecules contact the mem-brane on the pure solvent side Thus more water molecules cross the mem-brane from the pure solvent side than from the solution side

The additional water molecules on the solution side of the membrane cre-ate pressure and push some water molecules back across the membrane Theamount of additional pressure caused by the water molecules that moved intothe solution is called the osmotic pressure Osmotic pressure depends uponthe number of solute particles in a given volume of solution Thereforeosmotic pressure is another colligative property of solutions

PRACTICE PROBLEMS33 What are the boiling point and freezing point of a 0625m aqueous

solution of any nonvolatile nonelectrolyte solute

34 What are the boiling point and freezing point of a 040m solution ofsucrose in ethanol

35 A lab technician determines the boiling point elevation of an aque-ous solution of a nonvolatile nonelectrolyte to be 112degC What is thesolutionrsquos molality

36 A student dissolves 0500 mol of a nonvolatile nonelectrolyte solute inone kilogram of benzene (C6H6) What is the boiling point elevationof the resulting solution

For more practice withboiling point elevationand freezing pointdepression problems

go to SupplementalPractice Problems inAppendix A

Practice

Renal DialysisTechnicianPeople with poorly functioningkidneys must routinely undergodialysis in order to survive Theprocess which removes impuri-ties from the blood throughosmosis can be stressful for thepatient If you enjoy workingwith and helping people andwant to work in a medical set-ting consider being a renaldialysis technician

Dialysis technicians preparethe dialysis equipment and thepatient for treatment start thetreatment and monitor theprocess They also make neces-sary adjustments keep recordsand respond to emergenciesTechnicians spend more timewith the patients than thedoctors or nurses They workin hospitals outpatient facili-ties and home-based dialysisprograms

chemistrymccomself_check_quiz

476 Chapter 15 Solutions

Section 154 Heterogeneous Mixtures

Objectivesbull Identify the properties of

suspensions and colloids

bull Describe different types ofcolloids

bull Explain the electrostaticforces in colloids

VocabularysuspensioncolloidBrownian motionTyndall effect

As you learned in Chapter 3 most of the forms of matter that you encounterare mixtures A mixture is a combination of two or more substances that keeptheir basic identity Components of a mixture come in contact with each otherbut do not undergo chemical change You have been studying homogeneousmixtures called solutions so far in this chapter Not all mixtures are solutionshowever Heterogeneous mixtures contain substances that exist in distinctphases Two types of heterogeneous mixtures are suspensions and colloids

SuspensionsLook at the mixture shown in Figure 15-19a Although it resembles milkit is actually a freshly-made mixture of cornstarch stirred into water If youlet this mixture stand undisturbed for a while it separates into two distinctlayers a thick white pastelike substance on the bottom and the water ontop Cornstarch in water is a suspension a mixture containing particles thatsettle out if left undisturbed As shown in Figure 15-19b pouring a liquidsuspension through a filter also separates out the suspended particles Otherexamples of suspensions include fine sand in water and muddy water

Suspended particles are large compared to solvated particles with diame-ters greater than 1000 nm (106 m) compared to diameters less than 1 nm(19 m) for solvated particles Gravity acts on suspended particles in a shorttime causing them to settle out of the mixture Interestingly the settled outcornstarch particles form a solidlike state on the bottom of the containerHowever when stirred the solidlike state quickly begins flowing like a liq-uid Substances that behave this way are called thixotropic One of the mostcommon applications of a thixotropic mixture is house paint The paint flowsrather easily when applied with a brush but quickly thickens to a solidlikestate The quickly thickening paint helps it stick to the house and preventsruns in the paint

Figure 15-19

A suspension is a type of hetero-geneous mixture Suspensionparticles settle out over time and can be separated from themixture by filtration b

a

a b

Chem MC-476

null

14753836

CMC_Spn_C15-S04

null

17747762

ColloidsParticles in a suspension are much larger than atoms In contrast particles ina solution are atomic-scale in size A heterogeneous mixture of intermediatesize particles (between the size of solution particles and suspension particles)is a colloid Colloid particles are between 1 nm and 1000 nm in diameter Themost abundant substance in the mixture is the dispersion medium Milk is acolloid Although homogenized milk resembles the cornstarch mixture inFigure 15-19a you cannot separate its components by settling or by filtration

Colloids are categorized according to the phases of their dispersed parti-cles and dispersing mediums Milk is a colloidal emulsion because liquid par-ticles are dispersed in a liquid medium Other types of colloids are describedin Table 15-6 and shown in Figure 15-20 How many of them are familiarto you Can you name others

154 Heterogeneous Mixtures 477

Types of Colloids

Dispersed DispersingCategory particles medium Example

Solid sol Solid Solid Colored gems

Sol Solid Liquid Blood gelatin

Solid emulsion Liquid Solid Butter cheese

Emulsion Liquid Liquid Milk mayonnaise

Solid foam Gas Solid Marshmallow soaps that float

Foam Gas Liquid Whipped cream beaten egg white

Aerosol Solid Gas Smoke dust in air

Aerosol Liquid Gas Spray deodorant clouds

Table 15-6

Figure 15-20

All of the photos shown hereare examples of colloids

Fog which is a liquidaerosol is formed when smallliquid particles are dispersed in agas Many paints are sols afluid colloidal system of finesolid particles in a liquidmedium The bagel has acoarse texure with lots of smallholes throughout These smallholes are formed by a type offoam produced by yeast in thedough

c

b

a

a b c

LAB

See page 959 in Appendix E forIdentifying Colloids

Chem MC-477

null

6290336

Brownian motion If you were to observe a liquid colloid under the mag-nification of a microscope you would see that the dispersed particles makejerky random movements This erratic movement of colloid particles is calledBrownian motion It was first observed by and later named for the Scottishbotanist Robert Brown who noticed the random movements of pollen grainsdispersed in water Brownian motion results from collisions of particles of

the dispersion medium with the dispersed particles Thesecollisions prevent the colloid particles from settling out ofthe mixture

The reason the dispersed particles do not settle out isbecause they have polar or charged atomic groups on theirsurfaces These areas on their surfaces attract the posi-tively or negatively charged areas of the dispersing-medium particles This results in the formation ofelectrostatic layers around the particles as shown inFigure 15-21 The layers repel each other when the dis-persed particles collide and thus the particles remain inthe colloid If you interfere with the electrostatic layeringcolloid particles will settle out of the mixture For exam-ple if you stir an electrolyte into a colloid the dispersedparticles increase in size destroying the colloid Heatingalso destroys a colloid because it gives colliding particlesenough kinetic energy to overcome the electrostatic forcesand settle out

478 Chapter 15 Solutions

problem-solving LAB

How can you measure the tur-bidity of a colloidDesigning an Experiment The presence ofcloudy water or turbidity in a lake or river resultsfrom the suspension of solids in the water Most ofthese colloid particles come from erosion indus-trial and human waste algae blooms from fertiliz-ers and decaying organic matter The degree ofturbidity can greatly affect the health of a body ofwater in a number of ways but its greatest impactis that it prevents light from reaching vegetation

AnalysisThe Tyndall effect can be used to measure the tur-bidity of water Your goal is to plan a procedureand develop a scale to interpret the data

Thinking CriticallyYour plan should address the following 1 What materials would you need to measure

the Tyndall effect Can you select CBLs computers or graphing calculators to collector interpret data

2 What variables can be used to relate the abil-ity of light to pass through the liquid and thenumber of the colloid particles present

3 How would you change the turbidity in theexperiment

4 What variable would change and by how much

5 What will you use as a control

6 How could you relate the variables used in theexperiment to the actual number of colloidparticles that are present

7 Are there any safety precautions you mustconsider

Colloidalparticle

Repulsion

Attraction

Figure 15-21

The dispersing medium particlesform charged layers around thecolloid particles These chargedlayers repel each other and keepthe particles from settling out

Chem MC-478

null

8923362

The Tyndall effect Whereas concentrated colloids are often cloudy oropaque dilute colloids sometimes appear as clear as solutions Dilute colloidsappear to be homogeneous because their dispersed particles are so small thatthey cannot be seen by the unaided eye However dispersed colloid particlesare large enough to scatter light a phenomenon known as the Tyndall effectIn Figure 15-22b a beam of light is shone through two unknown mixturesWhich mixture is the colloid Which is the solution You can see that dis-persed colloid particles scatter the light unlike solvated particles in the solu-tion Solutions never exhibit the Tyndall effect Suspensions such as thecornstarch shown in Figure 15-19a also exhibit the Tyndall effect You haveobserved the Tyndall effect if you have observed rays of sunlight passingthrough smoke-filled air or viewed lights through fog at night as shown inFigure 15-22a Do the problem-solving LAB on the previous page to seehow the Tyndall effect can be used to determine the amount of colloid parti-cles in suspension

154 Heterogeneous Mixtures 479

Section 154 Assessment

42 Distinguish between suspensions and colloids

43 Describe different types of colloids

44 Why do dispersed colloid particles stay dispersed

45 Thinking Critically Use the Tyndall effect toexplain why it is more difficult to drive throughfog using high beams than using low beams

46 Comparing and Contrasting Make a table thatcompares the properties of solutions suspensionsand colloids

Figure 15-22

Two examples of the Tyndalleffect are shown here Thescattering of light by fog greatlyreduces visibility The beamof light is visible in the colloidbecause of light scattering

b

a

a b

chemistrymccomself_check_quiz

ColloidsParticles in a suspension are much larger than atoms In contrast particles ina solution are atomic-scale in size A heterogeneous mixture of intermediatesize particles (between the size of solution particles and suspension particles)is a colloid Colloid particles are between 1 nm and 1000 nm in diameter Themost abundant substance in the mixture is the dispersion medium Milk is acolloid Although homogenized milk resembles the cornstarch mixture inFigure 15-19a you cannot separate its components by settling or by filtration

Colloids are categorized according to the phases of their dispersed parti-cles and dispersing mediums Milk is a colloidal emulsion because liquid par-ticles are dispersed in a liquid medium Other types of colloids are describedin Table 15-6 and shown in Figure 15-20 How many of them are familiarto you Can you name others

154 Heterogeneous Mixtures 477

Types of Colloids

Dispersed DispersingCategory particles medium Example

Solid sol Solid Solid Colored gems

Sol Solid Liquid Blood gelatin

Solid emulsion Liquid Solid Butter cheese

Emulsion Liquid Liquid Milk mayonnaise

Solid foam Gas Solid Marshmallow soaps that float

Foam Gas Liquid Whipped cream beaten egg white

Aerosol Solid Gas Smoke dust in air

Aerosol Liquid Gas Spray deodorant clouds

Table 15-6

Figure 15-20

All of the photos shown hereare examples of colloids

Fog which is a liquidaerosol is formed when smallliquid particles are dispersed in agas Many paints are sols afluid colloidal system of finesolid particles in a liquidmedium The bagel has acoarse texure with lots of smallholes throughout These smallholes are formed by a type offoam produced by yeast in thedough

c

b

a

a b c

LAB

See page 959 in Appendix E forIdentifying Colloids

Chem MC-477

null

6290336

Brownian motion If you were to observe a liquid colloid under the mag-nification of a microscope you would see that the dispersed particles makejerky random movements This erratic movement of colloid particles is calledBrownian motion It was first observed by and later named for the Scottishbotanist Robert Brown who noticed the random movements of pollen grainsdispersed in water Brownian motion results from collisions of particles of

the dispersion medium with the dispersed particles Thesecollisions prevent the colloid particles from settling out ofthe mixture

The reason the dispersed particles do not settle out isbecause they have polar or charged atomic groups on theirsurfaces These areas on their surfaces attract the posi-tively or negatively charged areas of the dispersing-medium particles This results in the formation ofelectrostatic layers around the particles as shown inFigure 15-21 The layers repel each other when the dis-persed particles collide and thus the particles remain inthe colloid If you interfere with the electrostatic layeringcolloid particles will settle out of the mixture For exam-ple if you stir an electrolyte into a colloid the dispersedparticles increase in size destroying the colloid Heatingalso destroys a colloid because it gives colliding particlesenough kinetic energy to overcome the electrostatic forcesand settle out

478 Chapter 15 Solutions

problem-solving LAB

How can you measure the tur-bidity of a colloidDesigning an Experiment The presence ofcloudy water or turbidity in a lake or river resultsfrom the suspension of solids in the water Most ofthese colloid particles come from erosion indus-trial and human waste algae blooms from fertiliz-ers and decaying organic matter The degree ofturbidity can greatly affect the health of a body ofwater in a number of ways but its greatest impactis that it prevents light from reaching vegetation

AnalysisThe Tyndall effect can be used to measure the tur-bidity of water Your goal is to plan a procedureand develop a scale to interpret the data

Thinking CriticallyYour plan should address the following 1 What materials would you need to measure

the Tyndall effect Can you select CBLs computers or graphing calculators to collector interpret data

2 What variables can be used to relate the abil-ity of light to pass through the liquid and thenumber of the colloid particles present

3 How would you change the turbidity in theexperiment

4 What variable would change and by how much

5 What will you use as a control

6 How could you relate the variables used in theexperiment to the actual number of colloidparticles that are present

7 Are there any safety precautions you mustconsider

Colloidalparticle

Repulsion

Attraction

Figure 15-21

The dispersing medium particlesform charged layers around thecolloid particles These chargedlayers repel each other and keepthe particles from settling out

Chem MC-478

null

8923362

The Tyndall effect Whereas concentrated colloids are often cloudy oropaque dilute colloids sometimes appear as clear as solutions Dilute colloidsappear to be homogeneous because their dispersed particles are so small thatthey cannot be seen by the unaided eye However dispersed colloid particlesare large enough to scatter light a phenomenon known as the Tyndall effectIn Figure 15-22b a beam of light is shone through two unknown mixturesWhich mixture is the colloid Which is the solution You can see that dis-persed colloid particles scatter the light unlike solvated particles in the solu-tion Solutions never exhibit the Tyndall effect Suspensions such as thecornstarch shown in Figure 15-19a also exhibit the Tyndall effect You haveobserved the Tyndall effect if you have observed rays of sunlight passingthrough smoke-filled air or viewed lights through fog at night as shown inFigure 15-22a Do the problem-solving LAB on the previous page to seehow the Tyndall effect can be used to determine the amount of colloid parti-cles in suspension

154 Heterogeneous Mixtures 479

Section 154 Assessment

42 Distinguish between suspensions and colloids

43 Describe different types of colloids

44 Why do dispersed colloid particles stay dispersed

45 Thinking Critically Use the Tyndall effect toexplain why it is more difficult to drive throughfog using high beams than using low beams

46 Comparing and Contrasting Make a table thatcompares the properties of solutions suspensionsand colloids

Figure 15-22

Two examples of the Tyndalleffect are shown here Thescattering of light by fog greatlyreduces visibility The beamof light is visible in the colloidbecause of light scattering

b

a

a b

chemistrymccomself_check_quiz

Brownian motion If you were to observe a liquid colloid under the mag-nification of a microscope you would see that the dispersed particles makejerky random movements This erratic movement of colloid particles is calledBrownian motion It was first observed by and later named for the Scottishbotanist Robert Brown who noticed the random movements of pollen grainsdispersed in water Brownian motion results from collisions of particles of

the dispersion medium with the dispersed particles Thesecollisions prevent the colloid particles from settling out ofthe mixture

The reason the dispersed particles do not settle out isbecause they have polar or charged atomic groups on theirsurfaces These areas on their surfaces attract the posi-tively or negatively charged areas of the dispersing-medium particles This results in the formation ofelectrostatic layers around the particles as shown inFigure 15-21 The layers repel each other when the dis-persed particles collide and thus the particles remain inthe colloid If you interfere with the electrostatic layeringcolloid particles will settle out of the mixture For exam-ple if you stir an electrolyte into a colloid the dispersedparticles increase in size destroying the colloid Heatingalso destroys a colloid because it gives colliding particlesenough kinetic energy to overcome the electrostatic forcesand settle out

478 Chapter 15 Solutions

problem-solving LAB

How can you measure the tur-bidity of a colloidDesigning an Experiment The presence ofcloudy water or turbidity in a lake or river resultsfrom the suspension of solids in the water Most ofthese colloid particles come from erosion indus-trial and human waste algae blooms from fertiliz-ers and decaying organic matter The degree ofturbidity can greatly affect the health of a body ofwater in a number of ways but its greatest impactis that it prevents light from reaching vegetation

AnalysisThe Tyndall effect can be used to measure the tur-bidity of water Your goal is to plan a procedureand develop a scale to interpret the data

Thinking CriticallyYour plan should address the following 1 What materials would you need to measure

the Tyndall effect Can you select CBLs computers or graphing calculators to collector interpret data

2 What variables can be used to relate the abil-ity of light to pass through the liquid and thenumber of the colloid particles present

3 How would you change the turbidity in theexperiment

4 What variable would change and by how much

5 What will you use as a control

6 How could you relate the variables used in theexperiment to the actual number of colloidparticles that are present

7 Are there any safety precautions you mustconsider

Colloidalparticle

Repulsion

Attraction

Figure 15-21

The dispersing medium particlesform charged layers around thecolloid particles These chargedlayers repel each other and keepthe particles from settling out

Chem MC-478

null

8923362

The Tyndall effect Whereas concentrated colloids are often cloudy oropaque dilute colloids sometimes appear as clear as solutions Dilute colloidsappear to be homogeneous because their dispersed particles are so small thatthey cannot be seen by the unaided eye However dispersed colloid particlesare large enough to scatter light a phenomenon known as the Tyndall effectIn Figure 15-22b a beam of light is shone through two unknown mixturesWhich mixture is the colloid Which is the solution You can see that dis-persed colloid particles scatter the light unlike solvated particles in the solu-tion Solutions never exhibit the Tyndall effect Suspensions such as thecornstarch shown in Figure 15-19a also exhibit the Tyndall effect You haveobserved the Tyndall effect if you have observed rays of sunlight passingthrough smoke-filled air or viewed lights through fog at night as shown inFigure 15-22a Do the problem-solving LAB on the previous page to seehow the Tyndall effect can be used to determine the amount of colloid parti-cles in suspension

154 Heterogeneous Mixtures 479

Section 154 Assessment

42 Distinguish between suspensions and colloids

43 Describe different types of colloids

44 Why do dispersed colloid particles stay dispersed

45 Thinking Critically Use the Tyndall effect toexplain why it is more difficult to drive throughfog using high beams than using low beams

46 Comparing and Contrasting Make a table thatcompares the properties of solutions suspensionsand colloids

Figure 15-22

Two examples of the Tyndalleffect are shown here Thescattering of light by fog greatlyreduces visibility The beamof light is visible in the colloidbecause of light scattering

b

a

a b

chemistrymccomself_check_quiz

The Tyndall effect Whereas concentrated colloids are often cloudy oropaque dilute colloids sometimes appear as clear as solutions Dilute colloidsappear to be homogeneous because their dispersed particles are so small thatthey cannot be seen by the unaided eye However dispersed colloid particlesare large enough to scatter light a phenomenon known as the Tyndall effectIn Figure 15-22b a beam of light is shone through two unknown mixturesWhich mixture is the colloid Which is the solution You can see that dis-persed colloid particles scatter the light unlike solvated particles in the solu-tion Solutions never exhibit the Tyndall effect Suspensions such as thecornstarch shown in Figure 15-19a also exhibit the Tyndall effect You haveobserved the Tyndall effect if you have observed rays of sunlight passingthrough smoke-filled air or viewed lights through fog at night as shown inFigure 15-22a Do the problem-solving LAB on the previous page to seehow the Tyndall effect can be used to determine the amount of colloid parti-cles in suspension

154 Heterogeneous Mixtures 479

Section 154 Assessment

42 Distinguish between suspensions and colloids

43 Describe different types of colloids

44 Why do dispersed colloid particles stay dispersed

45 Thinking Critically Use the Tyndall effect toexplain why it is more difficult to drive throughfog using high beams than using low beams

46 Comparing and Contrasting Make a table thatcompares the properties of solutions suspensionsand colloids

Figure 15-22

Two examples of the Tyndalleffect are shown here Thescattering of light by fog greatlyreduces visibility The beamof light is visible in the colloidbecause of light scattering

b

a

a b

chemistrymccomself_check_quiz

480 Chapter 15 Solutions

Pre-Lab

1 Read the entire CHEMLAB procedure

2 What is the total volume of solution in each testtube Calculate the percent by volume of the solu-tions in test tubes 1 through 5 Prepare a data table

3 Review with your teacher how a colorimeterworks How are absorbance (A) and transmittance(T ) related

4 What is occurring during step 3 of the procedureWhy is a cuvette of water used

Procedure

1 Transfer 30 mL of blue dye stock solution into abeaker Transfer 30 mL of distilled water intoanother beaker

2 Label five clean dry test tubes 1 through 5

3 Pipette 2 mL of blue dye stock solution from thebeaker into test tube 1 4 mL into test tube 2 6 mLinto test tube 3 and 8 mL into test tube 4CAUTION Always pipette using a pipette bulb

4 With another pipette transfer 8 mL of distilledwater from the beaker into test tube 1 6 mL intotube 2 4 mL into tube 3 and 2 mL into tube 4

5 Mix the solution in test tube 1 with a stirring rodRinse and dry the stirring rod Repeat this proce-dure with each test tube

Safety Precautions

bull Always wear safety goggles and a lab apronbull The food-coloring solution can stain clothes

ProblemHow is light absorbance usedto find the concentrationof a blue dye solution

Objectivesbull Prepare solutions of known

concentration from a bluedye stock solution

bull Measure the absorbance ofknown and unknown aque-ous solutions

bull Infer the relationshipbetween light absorbanceand concentration of asolution

MaterialsCBL systemgraphing calculatorVernier colorimeterDIN adaptor and

cableTI Graph-Link

(optional)cuvettecotton swabstissues for wiping

cuvette

blue dye stocksolution

unknown solutiondistilled water50-mL graduated

cylinder100-mL beaker (2)small test tube (5)test tube rackpipette (2)pipette bulbstirring rod

Beerrsquos LawFinding the concentration of an unknown solution is an important

procedure in laboratory work One method commonly used todetermine solution concentration is to measure how much of a singlewavelength of light is absorbed by the solution and compare it toknown values of concentration and wavelength Light absorbance isdirectly related to the concentration of a solution This relationship iscalled Beerrsquos law

CHEMLAB CBL15

Known and Unknown Solutions Data

Test tube Concentration () T A

1

2

3

4

5

Unknown

CHEMLAB 481

6 Pipette 10 mL of blue dye stock into test tube 5

7 Load the ChemBio program into the calculatorConnect the CBL to the colorimeter using a DINadaptor Connect the CBL to the calculator usinga link cable Begin the ChemBio program on thecalculator Select 1 probe Select 4 COL-ORIMETER Enter Channel 1

8 Fill a cuvette about three-fourths full with dis-tilled water and dry its outside with a tissue Tocalibrate the colorimeter place the cuvette in thecolorimeter and close the lid Turn the wave-length knob to 0T Press TRIGGER on theCBL and enter 0 into the calculator Turn thewavelength knob to Red (635 nm) Press TRIG-GER on the CBL and enter 100 into the calcula-tor Leave the colorimeter set on Red for the restof the lab Remove the cuvette from the col-orimeter Empty the distilled water from thecuvette Dry the inside of the cuvette with a cleancotton swab

9 Select COLLECT DATA from the MAIN MENUSelect TRIGGERPROMPT from the DATACOLLECTION menu Fill the cuvette about three-fourths full with the solution from test tube 1 Dry the outside of the cuvette with a tissue andplace the cuvette in the colorimeter Close the lidAfter 10 to 15 seconds press TRIGGER andenter the concentration in percent from your datatable into the calculator Remove the cuvette andpour out the solution Rinse the inside of thecuvette with distilled water and dry it with aclean cotton swab Repeat this step for test tubes2 through 5

10 Select STOP AND GRAPH from the DATACOLLECTION menu when you have finishedwith data collection Draw the graph or use theTI Graph-Link to make a copy of the graph fromthe calculator screen You also will want to copythe data from the STAT list into your data table(or you can print it from a screen print usingGraph-Link)

11 Clean the cuvette with a cotton swab and fill itabout three-fourths full with the unknown dyesolution Place the cuvette in the colorimeter andclose the lid From the MAIN MENU selectCOLLECT DATA (do not select SET UPPROBES as this will erase your data lists) Select

MONITOR INPUT from the DATA COLLEC-TION MENU Press ENTER to monitor theabsorbance value of the colorimeter After about10-15 seconds record the absorbance value andrecord it in your data table

Cleanup and Disposal

1 All of the blue dye solutions can be rinsed downthe drain

2 Turn off the colorimeter Clean and dry thecuvette Return all equipment to its proper place

Analyze and Conclude

1 Analyzing Data Evaluate how close your graphis to the direct relationship exhibited by Beerrsquoslaw by doing a linear-regression line Select FITCURVE from the MAIN MENU (do not selectSET UP PROBES as this will erase your datalists) Select LINEAR L1L2 The calculator willgive you an equation in the form of y ax b One indicator of the fit of your graphis the size of b A small value of b means thegraph passes close to the origin The closer thecorrelation coefficient r reported by the programis to 100 the better the fit of the graph

2 Drawing Conclusions Use the graph of yourabsorbance and concentration data to determinethe concentration of your unknown solution

3 Form a Hypothesis Would you obtain the samedata if red dye was used Explain

4 Analyze your b and r valuesHow closely do your results match Beerrsquos lawReexamine the procedure and suggest reasonswhy the correlation coefficient from your datadoes not equal 100

Real-World Chemistry

1 Explain how Beerrsquos law can be applied in fooddrug and medical testing

2 The reaction of alcohol with orange dichromateions to produce blue-green chromium(III) ions isused in the Breathalyzer test a test that measuresthe presence of alcohol in a personrsquos breath Howcould a colorimeter be used in this analysis

Error Analysis

CHAPTER 15 CHEMLAB

There are millions of red blood cells in a singledrop of blood Red blood cells play a crucial roletransporting oxygen throughout the body

Hemoglobin in Red Blood CellsThe cells pictured at the right appear red becausethey contain a bright red moleculemdasha protein calledhemoglobin One molecule of hemoglobin containsfour iron(II) ions Four oxygen molecules bind toeach molecule of hemoglobin as blood passesthrough the lungs As the blood circulates throughthe body hemoglobin releases oxygen to the cellsthat make up various body tissues

Healthy red blood cells are round and flexiblepassing easily through narrow capillaries Buthemoglobin in people with sickle cell disease dif-fers from normal hemoglobin

What makes sickle cells differentEach hemoglobin molecule consists of four chainsThe entire molecule contains 574 amino acidgroups In people with sickle cell disease two non-polar amino acid groups are substituted for twopolar amino acid groups As a result there is lessrepulsion between hemoglobin molecules whichallows abnormal hemoglobin molecules to clumptogether This causes the abnormal hemoglobin mol-ecules to be less soluble in the red blood cells espe-cially after oxygen has been absorbed through thelungs The abnormal hemoglobin forces the redblood cells to become rigid and C-shaped resem-bling the farming tool called a sickle

These unusually shaped sickle cells clog the cir-culatory system reducing blood flow Thereforeoxygen supply to nearby tissues is reduced Sicklecell disease causes pain anemia stroke and sus-ceptibility to infection

Searching for a CureWhile most individuals with sickle cell diseasehave Mediterranean or African ancestry it is com-mon practice to screen all newborns in the UnitedStates for the condition Research shows that earlyintervention can reduce the risk of serious infectionthe leading cause of death in children with sicklecell disease Intensive chemotherapy and stem cell

transfusions are being studied as ways to treat andpossibly cure sickle cell disease

1 Communicating Ideas Learn about the lat-est advances in sickle cell disease researchCreate a chart that shows major symptomstheir causes and their treatment Share yourfindings with other students in your class

2 Debating the Issue The intensive chemo-therapy required to treat sickle cell diseasedrastically weakens the immune system leav-ing the patient vulnerable to overwhelminginfection Five to eight percent of the childrenwho undergo this treatment do not surviveShould doctors be allowed to use this proce-dure to treat sickle cell disease

Investigating the Issue

Sickle Cell Disease

CHEMISTRY and

Society

Visit the Chemistry Web site atchemistrymccom to find links to moreinformation about sickle cell disease

482 Chapter 15 Solutions

CHEMLAB 481

6 Pipette 10 mL of blue dye stock into test tube 5

7 Load the ChemBio program into the calculatorConnect the CBL to the colorimeter using a DINadaptor Connect the CBL to the calculator usinga link cable Begin the ChemBio program on thecalculator Select 1 probe Select 4 COL-ORIMETER Enter Channel 1

8 Fill a cuvette about three-fourths full with dis-tilled water and dry its outside with a tissue Tocalibrate the colorimeter place the cuvette in thecolorimeter and close the lid Turn the wave-length knob to 0T Press TRIGGER on theCBL and enter 0 into the calculator Turn thewavelength knob to Red (635 nm) Press TRIG-GER on the CBL and enter 100 into the calcula-tor Leave the colorimeter set on Red for the restof the lab Remove the cuvette from the col-orimeter Empty the distilled water from thecuvette Dry the inside of the cuvette with a cleancotton swab

9 Select COLLECT DATA from the MAIN MENUSelect TRIGGERPROMPT from the DATACOLLECTION menu Fill the cuvette about three-fourths full with the solution from test tube 1 Dry the outside of the cuvette with a tissue andplace the cuvette in the colorimeter Close the lidAfter 10 to 15 seconds press TRIGGER andenter the concentration in percent from your datatable into the calculator Remove the cuvette andpour out the solution Rinse the inside of thecuvette with distilled water and dry it with aclean cotton swab Repeat this step for test tubes2 through 5

10 Select STOP AND GRAPH from the DATACOLLECTION menu when you have finishedwith data collection Draw the graph or use theTI Graph-Link to make a copy of the graph fromthe calculator screen You also will want to copythe data from the STAT list into your data table(or you can print it from a screen print usingGraph-Link)

11 Clean the cuvette with a cotton swab and fill itabout three-fourths full with the unknown dyesolution Place the cuvette in the colorimeter andclose the lid From the MAIN MENU selectCOLLECT DATA (do not select SET UPPROBES as this will erase your data lists) Select

MONITOR INPUT from the DATA COLLEC-TION MENU Press ENTER to monitor theabsorbance value of the colorimeter After about10-15 seconds record the absorbance value andrecord it in your data table

Cleanup and Disposal

1 All of the blue dye solutions can be rinsed downthe drain

2 Turn off the colorimeter Clean and dry thecuvette Return all equipment to its proper place

Analyze and Conclude

1 Analyzing Data Evaluate how close your graphis to the direct relationship exhibited by Beerrsquoslaw by doing a linear-regression line Select FITCURVE from the MAIN MENU (do not selectSET UP PROBES as this will erase your datalists) Select LINEAR L1L2 The calculator willgive you an equation in the form of y ax b One indicator of the fit of your graphis the size of b A small value of b means thegraph passes close to the origin The closer thecorrelation coefficient r reported by the programis to 100 the better the fit of the graph

2 Drawing Conclusions Use the graph of yourabsorbance and concentration data to determinethe concentration of your unknown solution

3 Form a Hypothesis Would you obtain the samedata if red dye was used Explain

4 Analyze your b and r valuesHow closely do your results match Beerrsquos lawReexamine the procedure and suggest reasonswhy the correlation coefficient from your datadoes not equal 100

Real-World Chemistry

1 Explain how Beerrsquos law can be applied in fooddrug and medical testing

2 The reaction of alcohol with orange dichromateions to produce blue-green chromium(III) ions isused in the Breathalyzer test a test that measuresthe presence of alcohol in a personrsquos breath Howcould a colorimeter be used in this analysis

Error Analysis

CHAPTER 15 CHEMLAB

There are millions of red blood cells in a singledrop of blood Red blood cells play a crucial roletransporting oxygen throughout the body

Hemoglobin in Red Blood CellsThe cells pictured at the right appear red becausethey contain a bright red moleculemdasha protein calledhemoglobin One molecule of hemoglobin containsfour iron(II) ions Four oxygen molecules bind toeach molecule of hemoglobin as blood passesthrough the lungs As the blood circulates throughthe body hemoglobin releases oxygen to the cellsthat make up various body tissues

Healthy red blood cells are round and flexiblepassing easily through narrow capillaries Buthemoglobin in people with sickle cell disease dif-fers from normal hemoglobin

What makes sickle cells differentEach hemoglobin molecule consists of four chainsThe entire molecule contains 574 amino acidgroups In people with sickle cell disease two non-polar amino acid groups are substituted for twopolar amino acid groups As a result there is lessrepulsion between hemoglobin molecules whichallows abnormal hemoglobin molecules to clumptogether This causes the abnormal hemoglobin mol-ecules to be less soluble in the red blood cells espe-cially after oxygen has been absorbed through thelungs The abnormal hemoglobin forces the redblood cells to become rigid and C-shaped resem-bling the farming tool called a sickle

These unusually shaped sickle cells clog the cir-culatory system reducing blood flow Thereforeoxygen supply to nearby tissues is reduced Sicklecell disease causes pain anemia stroke and sus-ceptibility to infection

Searching for a CureWhile most individuals with sickle cell diseasehave Mediterranean or African ancestry it is com-mon practice to screen all newborns in the UnitedStates for the condition Research shows that earlyintervention can reduce the risk of serious infectionthe leading cause of death in children with sicklecell disease Intensive chemotherapy and stem cell

transfusions are being studied as ways to treat andpossibly cure sickle cell disease

1 Communicating Ideas Learn about the lat-est advances in sickle cell disease researchCreate a chart that shows major symptomstheir causes and their treatment Share yourfindings with other students in your class

2 Debating the Issue The intensive chemo-therapy required to treat sickle cell diseasedrastically weakens the immune system leav-ing the patient vulnerable to overwhelminginfection Five to eight percent of the childrenwho undergo this treatment do not surviveShould doctors be allowed to use this proce-dure to treat sickle cell disease

Investigating the Issue

Sickle Cell Disease

CHEMISTRY and

Society

Visit the Chemistry Web site atchemistrymccom to find links to moreinformation about sickle cell disease

482 Chapter 15 Solutions

There are millions of red blood cells in a singledrop of blood Red blood cells play a crucial roletransporting oxygen throughout the body

Hemoglobin in Red Blood CellsThe cells pictured at the right appear red becausethey contain a bright red moleculemdasha protein calledhemoglobin One molecule of hemoglobin containsfour iron(II) ions Four oxygen molecules bind toeach molecule of hemoglobin as blood passesthrough the lungs As the blood circulates throughthe body hemoglobin releases oxygen to the cellsthat make up various body tissues

Healthy red blood cells are round and flexiblepassing easily through narrow capillaries Buthemoglobin in people with sickle cell disease dif-fers from normal hemoglobin

What makes sickle cells differentEach hemoglobin molecule consists of four chainsThe entire molecule contains 574 amino acidgroups In people with sickle cell disease two non-polar amino acid groups are substituted for twopolar amino acid groups As a result there is lessrepulsion between hemoglobin molecules whichallows abnormal hemoglobin molecules to clumptogether This causes the abnormal hemoglobin mol-ecules to be less soluble in the red blood cells espe-cially after oxygen has been absorbed through thelungs The abnormal hemoglobin forces the redblood cells to become rigid and C-shaped resem-bling the farming tool called a sickle

These unusually shaped sickle cells clog the cir-culatory system reducing blood flow Thereforeoxygen supply to nearby tissues is reduced Sicklecell disease causes pain anemia stroke and sus-ceptibility to infection

Searching for a CureWhile most individuals with sickle cell diseasehave Mediterranean or African ancestry it is com-mon practice to screen all newborns in the UnitedStates for the condition Research shows that earlyintervention can reduce the risk of serious infectionthe leading cause of death in children with sicklecell disease Intensive chemotherapy and stem cell

transfusions are being studied as ways to treat andpossibly cure sickle cell disease

1 Communicating Ideas Learn about the lat-est advances in sickle cell disease researchCreate a chart that shows major symptomstheir causes and their treatment Share yourfindings with other students in your class

2 Debating the Issue The intensive chemo-therapy required to treat sickle cell diseasedrastically weakens the immune system leav-ing the patient vulnerable to overwhelminginfection Five to eight percent of the childrenwho undergo this treatment do not surviveShould doctors be allowed to use this proce-dure to treat sickle cell disease

Investigating the Issue

Sickle Cell Disease

CHEMISTRY and

Society

Visit the Chemistry Web site atchemistrymccom to find links to moreinformation about sickle cell disease

482 Chapter 15 Solutions

Study Guide 483

CHAPTER STUDY GUIDE15

Vocabulary

Key Equations and Relationships

Summary151 What are solutionsbull A solute dissolves in a solvent during a process

called solvation When the solvent is water theprocess also is called hydration

bull Every substance has a characteristic solubility in agiven solvent

bull Factors that affect solubility include the nature ofthe solute and solvent temperature and pressure

bull Henryrsquos law states that the solubility (S) of a gas ina liquid is directly proportional to the pressure (P)of the gas above the liquid at a given temperature

152 Solution Concentrationbull The concentration of a solution is a quantitative

measure of the amount of solute in a given amountof solvent or solution

bull Measures of concentration include mass and volumepercentages molarity molality and mole fraction

bull A dilute solution can be prepared from a more con-centrated standard stock solution

153 Colligative Properties of Solutionsbull Physical properties affected by the concentration of

the solute but not the nature of the solute are calledcolligative properties

bull Colligative properties of solutions include vaporpressure lowering boiling point elevation freezingpoint depression and osmotic pressure

154 Heterogeneous Mixturesbull One of the key differences between solutions col-

loids and suspensions is particle size

bull The random motion of colloidal dispersions due tomolecular collisions is called Brownian motion

bull The scattering of light by colloidal particles is calledthe Tyndall effect The Tyndall effect can be used todistinguish colloids from solutions

bull Henryrsquos law PS1

1

PS2

2

(p 460)

bull Percent by mass 100(p 463)

bull Percent by volume 100(p 464)

bull Molarity (M)

(p 464)

bull Molarity-volume relationship M1V1 M2V2

(p 467)

bull Molality (m)

(p 469)

bull Mole fractions XA nA

n

A

nB XB

nA

n

B

nB

(p 470)

bull Boiling point elevation Tb Kbm (p 472)

bull Freezing point depression Tf Kfm (p 473)

moles of solutekilogram of solvent

moles of soluteliters of solution

volume of solutevolume of solution

mass of solutemass of solution

bull boiling point elevation (p 472)bull Brownian motion (p 478)bull colligative property (p 471)bull colloid (p 477)bull concentration (p 462)bull freezing point depression

(p 473)bull heat of solution (p 457)bull Henryrsquos law (p 460)

bull immiscible (p 454)bull insoluble (p 454)bull miscible (p 454)bull molality (p 469)bull molarity (p 464)bull mole fraction (p 470)bull osmosis (p 475)bull osmotic pressure (p 475)bull saturated solution (p 458)

bull solubility (p 457)bull soluble (p 454)bull solvation (p 455)bull supersaturated solution (p 459)bull suspension (p 476)bull Tyndall effect (p 479)bull unsaturated solution (p 458)bull vapor pressure lowering (p 472)

chemistrymccomself_check_quiz

484 Chapter 15 Solutions

Go to the Chemistry Web site atchemistrymccom for additionalChapter 15 Assessment

Concept Mapping47 Complete the following concept map using the follow-

ing terms molarity mole fraction molality moles ofsolute

Mastering Concepts48 What is the difference between solute and solvent

(151)

49 What determines whether a solute will be soluble in agiven solvent (151)

50 Explain the difference between saturated and unsatu-rated solutions (151)

51 What does it mean if two liquids are said to be misci-ble (151)

52 What are three ways to increase the rate of solvation(151)

53 Why are gases less soluble at higher temperatures(151)

54 What is the difference between percent by mass andpercent by volume (152)

55 What is the difference between molarity and molality(152)

56 Explain on a particle basis why the vapor pressure of asolution is lower than a pure solvent (153)

57 How does a solute affect the boiling point of a solu-tion (153)

58 How does a solute affect the freezing point of a solu-tion (153)

59 Describe osmosis (154)

60 What is a colligative property (154)

61 What is a suspension and how does it differ from acolloid (154)

62 Name a colloid formed from a gas dispersed in a liq-uid (154)

63 How can the Tyndall effect be used to distinguishbetween a colloid and a solution Why (154)

Mastering Problems

Henryrsquos Law (151)64 The solubility of a gas in water is 022 gL at 200 kPa

of pressure What is the solubility when the pressure isincreased to 115 kPa

65 The solubility of a gas in water is 066 gL at 15 kPaof pressure What is the solubility when the pressure isincreased to 400 kPa

66 The solubility of a gas is 20 gL at 500 kPa of pres-sure How much gas will dissolve in 1 L at a pressureof 100 kPa

67 The solubility of a gas is 45 gL at a pressure of10 atm At what pressure will there be 45 g of gas in10 L of solution

68 The partial pressure of CO2 inside a bottle of soft drinkis 40 atm at 25degC The solubility of CO2 is 012 molLWhen the bottle is opened the partial pressure drops to30 104 atm What is the solubility of CO2 in theopen drink Express your answer in grams per liter

Percent Solutions (152)69 Calculate the percent by mass of 355 g NaCl dis-

solved in 88 g water

70 Calculate the percent by mass of benzene in a solutioncontaining 142 g of benzene in 280 g of carbon tetra-chloride

71 What is the percent by volume of 25 mL of methanolin 75 mL of water

72 A solution is made by adding 123 mol KCl to10000 g of water What is the percent by mass ofKCl in this solution

73 What mass of water must be added to 2550 g NaCl tomake a 1500 percent by mass aqueous solution

74 The label on a 250-mL stock bottle reads 215 alco-hol by volume What volume of alcohol does itcontain

75 A 140 percent by mass solution of potassium iodidedissolved in water has a density of 1208 gmL Howmany grams of KI are in 250 mL of the solution

CHAPTER ASSESSMENTCHAPTER ASSESSMENT15

1

3

42

Volumeof solution

Total molesin solution

Kilogramsof solvent

chemistrymccomself_check_quiz

Assessment 485

CHAPTER 15 ASSESSMENT

Molarity (152)76 What is the molarity of the following solutions

a 25 mol KCl in 10 L of solutionb 135 mol H2SO4 in 245 mL of solutionc 0875 mol of ammonia in 155 mL of solution

77 What is the molarity of the following solutions

a 096 g MgCl2 in 500 mL of solutionb 933 g Na2S in 450 mL solutionc 248 g CaF2 in 375 mL of solution

78 How many moles of solute are contained in the fol-lowing solutions

a 1525 mL 210M CaCl2b 125 mL 00500M Ba(OH)2c 531 mL 122M HCl

79 How many grams of solute are contained in the fol-lowing solutions

a 643 mL 00238M KOHb 142 mL 140M K2SO4c 7500 mL 0225M NH4OH

Molar Dilution (152)80 How many milliliters of 255M NaOH is needed to

make 125 mL 075M NaOH solution

81 How many milliliters of 0400M HBr solution can bemade from 500 mL of 800M HBr solution

82 What is the molarity of each resulting solution whenthe following mixtures are prepared

a 5000 mL H2O is added to 200 mL 600M HNO3b 300 mL 175M HCl is added to 800 mL 0450M

HCl

Molality and Mole Fraction (152)83 Calculate the molality of the following solutions

a 157 g NaCl in 1000 g H2Ob 200 g CaCl2 in 7000 g H2Oc 376 g NaOH in 0850 L H2O

84 Calculate the mole fraction of NaCl CaCl2 andNaOH in the solutions listed in the previous problem

85 What are the molality and mole fraction of solute in a355 percent by mass aqueous solution of formic acid(HCOOH)

Colligative Properties (153)86 Using the information in Tables 15-4 and 15-5 calcu-

late the freezing point and boiling point of 120 g ofglucose (C6H12O6) in 500 g H2O

87 Using the information in Tables 15-4 and 15-5 calcu-late the freezing point and boiling point of each of thefollowing solutions

a 275m NaOH in waterb 0586m of water in ethanolc 126m of naphthalene (C10H8) in benzene

88 A rock salt (NaCl) ice and water mixture is used tocool milk and cream to make homemade ice creamHow many grams of rock salt must be added to waterto lower the freezing point 100degC

89 Calculate the freezing point and boiling point of asolution that contains 554 g NaCl and 423 g KBr dis-solved in 7503 mL H2O

Mixed ReviewSharpen your problem-solving skills by answering thefollowing

90 If you prepared a saturated aqueous solution of potas-sium chloride at 25degC and then heated it to 50degCwould you describe the solution as unsaturated satu-rated or supersaturated Explain

91 Use the graph below to explain why a carbonated bev-erage does not go flat as quickly when it contains ice

92 Which of the following substances will be soluble inthe nonpolar solvent carbon tetrachloride (CCl4) Br2C6H14 NaNO3 HCl Explain

93 How many grams of calcium nitrate (Ca(NO3)2) wouldyou need to prepare 300 L of a 0500M solution

94 What would be the molality of the solution describedin the previous problem

0 20 40 6010 30 50 70

Solu

bili

ty (

mo

lL)

8

7

6

5

4

3

2

1

0

Temperature (C)

Solubility of CO2in Water at 1 atm

486 Chapter 15 Solutions

Thinking Critically95 Inferring Why not spread a nonelectrolyte on a road

to help ice melt

96 Using Scientific Diagrams Complete the diagrambelow using the following phrases solution separatedsolvent + solute separated solvent + separated solutesolvent + solute Is the process described exothermicor endothermic

97 Designing an Experiment You are given a sampleof a solid solute and three aqueous solutions contain-ing that solute How would you determine which solu-tion is saturated unsaturated and supersaturated

98 Using Graphs The following solubility data was col-lected in an experiment Plot a graph of the molarityof KI versus temperature What is the solubility of KIat 55degC

99 Comparing Which of the following solutions hasthe highest concentration Rank the solutions fromthe greatest to the smallest boiling point depressionExplain your answer

a 010 mol NaBr in 1000 mL solutionb 21 mol KOH in 100 L solutionc 12 mol KMnO4 in 300 L solution

Writing in Chemistry100 Investigate the total amount of salt used in the US

Construct a circle graph showing the different usesand amounts Discuss each of these areas in detailSalt was once used as a currency of high value Findout why this was the case

101 Look up the various electrolytes in the human bloodstream and discuss the importance of each

102 Research the contents of the tank scuba divers typi-cally use How does its composition differ from theair that you breathe What is the condition known asthe bends How is it treated

Cumulative ReviewRefresh your understanding of previous chapters byanswering the following

103 The radius of an argon atom is 94 pm Assuming theatom is spherical what is the volume of an argonatom in nm3 V = 43r3 (Chapter 2)

104 Identify which of the following molecules is polar(Chapter 9)

a SiH4b NO2c H2Sd NCl3

105 Name the following compounds (Chapter 8)

a NaBrb Pb(CH3COO)2c (NH4)2CO3

106 A 120-g sample of an element contains 594 1022

atoms What is the unknown element (Chapter 11)

107 Pure bismuth can be produced by the reaction of bis-muth oxide with carbon at high temperatures

2Bi2O3 3C rarr 4Bi 3CO2

How many moles of Bi2O3 reacted to produce 126moles of CO2 (Chapter 12)

108 A gaseous sample occupies 324 mL at 23degC and075 atm What volume will it occupy at STP(Chapter 14)

CHAPTER ASSESSMENT15

3

4

2

1

Ener

gy

Solubility of KI Data

Temperature (degC) Grams of KI per 1000 g solution

20 144

40 162

60 176

80 192

100 206

Table 15-7

chemistrymccomchapter_test

486 Chapter 15 Solutions

Thinking Critically95 Inferring Why not spread a nonelectrolyte on a road

to help ice melt

96 Using Scientific Diagrams Complete the diagrambelow using the following phrases solution separatedsolvent + solute separated solvent + separated solutesolvent + solute Is the process described exothermicor endothermic

97 Designing an Experiment You are given a sampleof a solid solute and three aqueous solutions contain-ing that solute How would you determine which solu-tion is saturated unsaturated and supersaturated

98 Using Graphs The following solubility data was col-lected in an experiment Plot a graph of the molarityof KI versus temperature What is the solubility of KIat 55degC

99 Comparing Which of the following solutions hasthe highest concentration Rank the solutions fromthe greatest to the smallest boiling point depressionExplain your answer

a 010 mol NaBr in 1000 mL solutionb 21 mol KOH in 100 L solutionc 12 mol KMnO4 in 300 L solution

Writing in Chemistry100 Investigate the total amount of salt used in the US

Construct a circle graph showing the different usesand amounts Discuss each of these areas in detailSalt was once used as a currency of high value Findout why this was the case

101 Look up the various electrolytes in the human bloodstream and discuss the importance of each

102 Research the contents of the tank scuba divers typi-cally use How does its composition differ from theair that you breathe What is the condition known asthe bends How is it treated

Cumulative ReviewRefresh your understanding of previous chapters byanswering the following

103 The radius of an argon atom is 94 pm Assuming theatom is spherical what is the volume of an argonatom in nm3 V = 43r3 (Chapter 2)

104 Identify which of the following molecules is polar(Chapter 9)

a SiH4b NO2c H2Sd NCl3

105 Name the following compounds (Chapter 8)

a NaBrb Pb(CH3COO)2c (NH4)2CO3

106 A 120-g sample of an element contains 594 1022

atoms What is the unknown element (Chapter 11)

107 Pure bismuth can be produced by the reaction of bis-muth oxide with carbon at high temperatures

2Bi2O3 3C rarr 4Bi 3CO2

How many moles of Bi2O3 reacted to produce 126moles of CO2 (Chapter 12)

108 A gaseous sample occupies 324 mL at 23degC and075 atm What volume will it occupy at STP(Chapter 14)

CHAPTER ASSESSMENT15

3

4

2

1

Ener

gy

Solubility of KI Data

Temperature (degC) Grams of KI per 1000 g solution

20 144

40 162

60 176

80 192

100 206

Table 15-7

chemistrymccomchapter_test

Standardized Test Practice 487

Use these questions and the test-taking tip to preparefor your standardized test

1 How much water must be added to 60 mL of a0050M stock solution to dilute it to 0020M

a 15 mL c 60 mLb 90 mL d 24 mL

2 At a pressure of 100 atm and a temperature of 20ordmC172 g CO2 will dissolve in 1 L of water How muchCO2 will dissolve if the pressure is raised to 135 atmand the temperature stays the same

a 232 gL c 0785 gLb 127 gL d 0431 gL

3 What is the molality of a solution containing 025 gof dichlorobenzene (C6H4Cl2) dissolved in 100 g ofcyclohexane (C6H12)

a 017 molkg c 0025 molkgb 0014 molkg d 000017 molkg

4 If 1 mole of each of the solutes listed below is dis-solved in 1 L of water which solute will have thegreatest effect on the vapor pressure of its respectivesolutiona KBr c MgCl2b C6H12O6 d CaSO4

5 What volume of a 0125M NiCl2 solution contains325 g NiCl2a 406 mL c 385 mLb 201 mL d 260 mL

Interpreting Graphs Use the graph to answer ques-tions 6ndash8

6 What is the volume of bromine (Br2) in 7000 L ofSolution 1

a 5563 mLb 8808 mLc 1803 mLd 2718 mL

7 How many grams of Br2 are in 5500 g of Solution 4

a 3560 gb 0084 98 gc 1151 gd 02628 g

8 Which of the following relationships is true

a 2 Concentration of solution 2 Concentration of solution 3

b 05 Concentration of solution 2 Concentration of solution 3

c Concentration of solution 2 025 Concentration of solution 3

d Concentration of solution 2 3 Concentration of solution 3

9 Which is NOT a colligative property

a boiling point elevationb freezing point depressionc vapor pressure increased osmotic pressure

10 How can colloids be distinguished from solutions

a dilute colloids have particles that can be seen withthe naked eye

b colloid particles are much smaller than solvatedparticles

c colloid particles that are dispersed will settle out ofthe mixture in time

d colloids will scatter light beams that are shonethrough them

STANDARDIZED TEST PRACTICECHAPTER 15

Take a Break If you have a chance to take abreak or get up from your desk during a test takeit Getting up and moving around will give youextra energy and help you clear your mind Duringyour stretch break think about something otherthan the test so yoursquoll be able to get back to thetest with a fresh start

Perc

ent

1 2 3 400000

01000

031

89

02000

03000

04000

05000

06000

07000

08000

09000

Solution number

Bromine (Br2) Concentration ofFour Aqueous Solutions

079

47

025

75

015

96

015

45

047

79

Percent by mass

Percent by volume

010

30

005

15

chemistrymccomstandardized_test