Classifications of Mixtures• Heterogeneous Mixtures—composed of

different types of phases of substances- ex: Fruit salad Granite

• Homogeneous Mixtures—the same throughout (substances have dissolved in one another)

- ex: Salt water Alloys (metal mixtures)

Types of Mixtures

• Solutions

• Suspensions

• Colloids

Solutions are:• A homogeneous

mixture of two or more substances in a single phase

• Composed of:1. Solvent- the

substance that does the dissolving

2. Solute- the substance that is being dissolved

Example: In sugar water, water is the solvent and sugar is the solute.

Suspensions• A mixture in which the particles are so

large that they settle out unless the mixture is constantly stirred or agitated.

• Ex. A jar of muddy water• Ex: Orange juice

Colloids• A mixture in which the particles are

intermediate in size between those in solutions and suspensions.

• Particles are suspended in mixtureEx. Jello, Paints, Milk, Mayonnaise, Fog,

Cheese

Tyndall Effect• Colloids appear to be homogeneous

mixtures (aka solutions) because the individual particles cannot be seen…however, they are not true solutions.

• The particles are large enough to scatter light.

• This effect is known as the Tyndall Effect.

Types of Solutions:***Type of solution is determined

by the solvent: • Gaseous-mixture of two or more

gases• Liquid- solvent is a liquid• Solid- solvent is a solid

Examples of:1. Gaseous- air, scuba tanks2. Liquid- tea, Kool-aid, cokes, salt

water3. Solid- metal alloys, amalgams

(dental fillings)

Some solutions conduct electricity:

• Electrolytes- a solution that conducts electricity as a result of the formation of ions in solution (examples: salt water, vinegar)

• Nonelectrolyte- a solution that does not conduct electricity because there is no formation of ions in solution (example: sugar in oil)

What solutions of electrolytes look like versus solutions that

are not electrolytes:

Electrolyte

Nonelectrolyte

Factors that Affect Rates of Dissolving Solids in Liquids:1. Surface Area- increasing

surface area increases the rate of dissolving

2. Agitation- stirring or shaking increases rate of dissolving

3. Heat- heating the solvent will increase the rate of dissolving

Solubility• The amount of a

substance that is dissolved at solution equilibrium in a specific amount of solvent at a specified temperature.

• Factors Affecting Solubility:1. Nature of Solute and Solvent-

“like dissolves like”- polar dissolves polar-nonpolar dissolves nonpolar

2. Pressure- changes in pressure have little affect on dissolving solids in liquids but an increase in pressure will increase the solubility of gases in liquids

3. Temperaturea) Increasing temp., decreases gas

solubilityb) Increasing temp., increases solid

solubility

3 Classifications of Solutions• Saturated- a solution that contains

the max. amount of dissolved solute at a given temp.

• Unsaturated- a solution that contains less solute than a saturated solution at a given temp.

• Supersaturated- a solution that contains more dissolved solute than a saturated solution at a given temp.

Solubility Curves:• Show how much solute can go

into solution with a given amount of solvent at different temperatures.

Solubility Curve:

Colligative Properties• A property that depends on the

number of solute particles but is independent of their nature

• 3 Colligative Properties1. Vapor Pressure Lowering2. Freezing Point Depression3. Boiling Point Elevation

• Vapor Pressure Lowering- the vapor pressure of a solvent containing a nonvolatile solute is lower than the vapor pressure of the pure solvent at any temp.

• Freezing Point Depression- the freezing point of a solvent containing a solute will be lower than the pure solvent

• Boiling Point Elevation- the boiling point of a solvent containing a solute will be higher than the pure solvent

Solution Concentration:• Concentration is a

measurement of the amount of solute in a given amount of solvent or solution

• Can be expressed qualitatively (using words) or quantitatively (using numbers)

• Qualitative Terms:1. Dilute- relatively small

amount of solute compared to solvent

2. Concentrated- relatively large amount of solute in a solvent

Quantitative Terms:

• Percent by Mass• Molarity• Molality

Percent by Mass• The number of grams of solute dissolved in 100

g of water

Percent by mass = mass solute x100 (mass solute + mass solvent)

Examples:• A solution of sodium chloride is prepared by

dissolving 5 g of salt in 550 g of water. What is the concentration as given by percent by mass?

Percent by mass = mass solute x100 (mass solute + mass solvent)

Percent by mass = __5__ x 100 = 0.9% (5 + 550)

Answer: 0.9%

Examples:• What is the percent by mass of a solution

prepared by dissolving 4 g of acetic acid in 35 g of water.

Percent by mass = mass solute x100 (mass solute + mass solvent)

Percent by mass = __4__ x 100 = 10.26% (4 + 35)

Answer: 10.26%

Molarity• Symbolized by M• Units of mol L• Describes how many moles of solute

are present per liter of solution

Examples:• What is the molarity of 3.5 L of solution that

contains 90 g of sodium chloride?

Molarity = mol L

90 g NaCl x 1 mol NaCl = 1.54 mol NaCl 58.44 g NaCl

Molarity = mol = 1.54 mol solute = 0.44 M

L 3.5 L solution

Answer: 0.44 M

Examples:• How many moles of HCl are present in 0.8 L of a 0.5 M HCl solution?

Molarity = 0.5 M

Molarity = mol L

0.5 = mol HCl 0.8

Answer: mol HCl = 0.4 mol

Examples:• How many grams of sodium chloride will be

required to make 555 mL of a 1.45 M solution?

Molarity = 1.45 M 555 mL = 0.555 L Molarity = mol

L 1.45 = mol NaCl

0.555mol NaCl = 0.80475 mol

0.80475 mol NaCl x 58.44 g 1 mol NaCl

47.03 g NaCl

More Examples:• How many liters of solution can be prepared

if 78.9 g of sodium chloride is used to make a 3.00 M solution?

78.9 g NaCl x 1 mol NaCl = 1.35 mol NaCl 58.44 g NaCl

Molarity = mol L

3 = 1.35 L

Answer: 0.45 L

More Examples:• What is the molarity of a solution that is

prepared by using 20 g of sodium hydroxide in enough water to make a 2 L solution?

20 g NaOH x 1 mol NaOH = 0.5 mol NaOH 39.997 g NaOH

Molarity = mol = 0.5 mol = 0.25 M L 2 L

Answer: 0.25 M

Dilution Problems• Use the equation:

M1V1 = M2V2 Where M1 = molarity 1 V1= volume 1

M2 = molarity 2 V2 = volume 2

Examples:• What is the molarity of a solution that is

made by diluting 50 mL of a 4.74 M solution to 250 mL?

M1V1 = M2V2

(4.74)(0.05L) = (M2)(0.25)M2 = 0.948 M

Answ: 0.948 M

Molality

Symbolized by m

Units are mol solute kg of solvent

Describes how many moles of solute are present per kg of solvent.

Examples:• A solution contains 17.1 g of sucrose

(C12H22O11) dissolved in 125 g of water. Find the molal concentration.

Molality = mol solute kg solvent

Solute: 17.1 g C12H22O11 x 1 mol C12H22O11 = 0.05 mol C12H22O11

342.3 g C12H22O11

Solvent: 125 g water = 0.125 kg

Molality = mol solute = 0.05 = 0.400 m kg solvent 0.125

Answer: 0.400 m

Examples:• How much iodine (in grams) must be added to

prepare a 0.480 m solution of iodine in carbon tetrachloride (CCl4) if 100 g of CCl4 is used?

Molality = mol solute kg solvent

Solvent: 100 g CCl4 = 0.100 kg CCl4 Molality = mol solute

kg solvent 0.480 = mol solute

0.100 kg Mol solute = 0.048 mol iodine • 0.048 mol I2 x 253.8 g I2 = 12.2 g

1 mol I2

Answer: mol solute = 12.2 g

More Examples:• What is the molality of a solution composed of

2.55 g of acetone (CH3)2CO, dissolved in 200 g of water?

Molality = mol solute kg solvent Solute: 2.55 g (CH3)2CO x 1 mol (CH3)2CO = 0.0439 mol (CH3)2CO

58.08 g (CH3)2COSolvent: 200 g water = 0.200 kg water Molality = mol solute kg solvent Molality = 0.0439 / 0.200 = 0.220 m

Answer: 0.220 m

More Examples:• What quantity, in grams, of methanol

(CH3OH) is required to prepare a 0.244 m solution in 400 g of water?

Molality = mol solute kg solvent

Solvent: 400 g water = 0.400 kg 0.244 = mol solute

0.400 kg mol solute = 0.0976 kg CH3OH

0.0976 mol CH3OH x 32.04 g CH3OH = 3.13 g 1 mol CH3OH

Answer: 3.13 g

• How many grams of AgNO3 are needed to prepare a 0.125 m solution in 250 mL of water? (1 mL of water = 1g and 1 L of water = 1 kg)

Molality = mol solute kg solvent

Solvent: 250 mL water = 0.250 L water0.125 = mol solute

0.250 kg mol solute = 0.03125 mol AgNO3

0.03125 mol AgNO3 x 169.87 g AgNO3 = 5.31 g AgNO3

1 mol AgNO3

Answ: 5.31 g AgNO3

Examples:• 2H3PO4 + 3Ca(OH)2 Ca3(PO4)2 +6 H2O

If 2 L of 4 M phosphoric acid is used, how many grams of water could be formed?

Molarity = mol solute 4 = mol phosphoric acid mol phos acid = 8 mol L solution 2 liters

8 mol H3PO4 6 mol H20 18.01532 g H2O = 432.368 g H2O 2 mol H3PO4 1 mol H2O

Example2H3PO4 + 3Ca(OH)2 Ca3(PO4)2 +6 H2O

If 6 g of calcium hydroxide is used, how many liters of a 5 M acid solution would be needed?Molarity = mol solute L solution5 M = mol H3PO4

L solution

**Must solve for mol H3PO4 so we can plug it into the equation above

6 g Ca(OH)2 1 mol Ca(OH)2 2 mol H3PO4 = 0.054 mol H3PO4

74.0932 g Ca(OH)2 3 mol Ca(OH)2

5 M = mol H3PO4 5 = 0.054 mol H3PO4 L solution = 0.011 liters L solution L solution

2H3PO4 + 3Ca(OH)2 Ca3(PO4)2 +6 H2O

When 200 mL of 6 M phosphoric acid is used, how many mL of a 3 M calcium hydroxide solution would be required?

Molarity = mol solute L solution6 = mol phosphoric acid 0.200 Lmol phosphoric acid = 1.2 mol H3PO4

Molarity = mol solute L solution 3 = mol CaOH L solution