TYPES OF SOLUTIONS AND METHODS TO

EXPRESS CONCENTRATION

The present topic types of solutions and methods

to express concentration was introduced in class

IX and Class XI

Class IX: Chapter 2

IS MATTER AROUND US PURE?

2.2 What is a Solution?

i. Mass percent%

ii. Volume percent%

iii. Mass by volume %

CLASS XI: SOME BASIC CONCEPTS OF CHEMISTRY

1.10.2 Reactions in Solutions

A majority of reactions in the laboratories are carried out in solutions. Therefore, it is

important to understand as how the amount of substance is expressed when it is

present in the solution. The concentration of a solution or the amount of substance

present in its given volume can be expressed in any of the following ways.

Mass per cent or weight per cent (w/w %)

Mole fraction

Molarity

Molality

SOLUTION

SOLUTION is the homogeneous mixture of two or more than two components. Most of the solutions are binary i.e. consists of

two components out of which that is present in the largest quantity is called solvent & one which is present in smaller quantity

called solute. SOLUTION= SOLUTE + SOLVENT

EXAMPLES:

(i) A solution of sugar in water is a solid in liquid solution. In this solution, sugar is the solute and water is the

solvent.

(ii) A solution of iodine in alcohol known as ‘tincture of iodine’, has iodine (solid)as the solute and alcohol

(liquid) as the solvent.

(iii) Aerated drinks like soda water etc., are gas in liquid solutions. These contain carbon dioxide (gas) as solute

and water (liquid) as solvent.

(iv) Air is a mixture of gas in gas. Air is a homogeneous mixture of a number of gases. Its two main constituents

are, oxygen (21%) and nitrogen (78%). The other gases are present in very small quantities.

PROPERTIES OF A SOLUTION

• A solution is a homogeneous mixture.

• The particles of a solution are smaller than 1 nm (10-9 metre) in

diameter. So, They cannot be seen by naked eyes.

• Because of very small particle size, they do not scatter a beam of light

passing through the solution. So, the path of light is not visible in a

solution.

• The solute particles cannot be separated from the mixture by the process

of filtration. The solute particles do not settle down when left

undisturbed, that is, a solution is stable

DIFFERENT TYPES OF SOLUTIONS

1.On the Basis of Water as Solvent

Based on the whether the solvent is water or not, solutions are of two types.

Aqueous solutions: These solutions have water as the solvent. Examples of such

solutions are sugar in water, carbon dioxide in water, etc.

Non-Aqueous Solutions: These solutions have a solvent that is not water. It

could be ether, benzene, petrol, carbon tetrachloride etc. Common examples

include sulfur in carbon disulphide, naphthalene in benzene, etc.

2. On the Basis of the Amount of Solute Added

Based on the amount of solute present in the solution, we can classify them into the following types.

Unsaturated Solution: An unsaturated is one that can dissolve more solute at a definite temperature. It means that we can still add more solute to the solvent.

Saturated Solution: A solution is said to be saturated when we can’t add any more solute to the solvent. This means that the solution can’t dissolve any more solute at a

definite temperature.

Supersaturated Solution: A supersaturated solution is one where the solute is present in an excess amount. This solute is dissolved forcefully by raising the temperature

or pressure of the solution. These generally crystal out in the bottom by the method called crystallisation.

3. On the Basis of Amount of Solvent Added

Concentrated Solution: A concentrated solution

has large amounts of solute in the given

solvent. Examples include Brine solution,

Orange juice, dark colour tea.

Dilute Solution: A dilute solution has a small

amount of solute in a large amount of solvent.

Examples include Salt solution, light colour tea.

4. On the basis of physical states of solute and

solvent

Type of Solution Solute Solvent Common Examples

Gaseous Solutions Gas Gas Mixture of oxygen and nitrogen

gases

Liquid Gas Chloroform mixed with nitrogen

gas

Solid Gas Camphor in nitrogen gas

Liquid Solutions Gas Liquid Oxygen dissolved in water

Liquid Liquid Ethanol dissolved in water

Solid Liquid Glucose dissolved in water

Solid Solutions Gas Solid Solution of hydrogen in

palladium

Liquid Solid Amalgam of mercury with

sodium

Solid Solid Copper dissolved in gold

CONCENTRATION: The amount of solute present in the given

quantity of solvent or solution

EXPRESSING CONCENTRATION OF SOLUTIONS

METHODS TO EXPRESS CONCENTRATION

i. Mass percentage :Mass of solute per100g of solution

Mass % = (mass of solute / total mass of solution) X 100

ii. Volume percentage : volume of soluteper100mlof solution

Volume % = (volume of solute/ total volume of

solution) X 100

iii.Mass by volume percentage (w/V): Another unit which is commonly

used in medicine and pharmacy is mass by volume percentage. It is the mass

of solute dissolved in 100 mL of the solution.

Mass by Volume % = (mass of solute/ total volume of

solution) X 100

iv. Parts per million: parts of a component per million parts of the solution.

Parts per million =(Number of parts of the component /Total number of

parts of all components of the solution) ×106

v. Mole fraction(x): It is the ratio of no. of moles of one component to

the total no. of all the components

present in the solution . For binary solution:‐the no. of

moles of A and B are nA and nB

respectively

so, xA=nA/nA+nB ;xB= nB/ nA+nB

In binary solution xA+ xB=1

vi. Molarity(M): No. of moles of solute dissolved in one litre of solution.

Molarity(M) = moles of solute/ vol. of solution in litre

SHORT FORM: MOLAR

vii. molality(m):No. of moles of solute per kg of the solvent.

molality(m) = moles of solute/mass of solvent in kg

SHORT FORM: MOLAL

Molality is independent of temp. whereas molarity is a function of temp.

because vol. depends on temp. and mass does not.

1.Calculate the mass percentage of aspirin (C9H8O4) in

acetonitrile (CH3CN) when 6.5 g of C9H8O4 is

dissolved in 450 g of CH3CN .

Ans. Mass of solution = 6.5g + 450g = 456.5g

Mass% of aspirin= Mass of aspirinX100

Mass of solution

=6.5/456.5 X 100 = 1.424%

NUMERICALS ON CONCENTRATION

2. Calculate the molarity of a solution containing 5 g of

NaOH in 450 mL.

Ans. Moles of NaOH =5 g/40 g /mol(molar mass of

NaOH=40 g /mol)

= 0.125 mol

Volume of the solution in litres = 450 mL / 1000mL/L

=0.45L

Molarity(M)= Moles of NaOH / Volume of the solution in litres

= 0.125 mol / 0.45L

= 0.278 mol /L

HOW TO PREPARE SOLUTIONS IN LABORATORY

Preparation of standard solution

What is standard solution?

(A solution whose concentration is known)

For example to prepare 1M Na OH solution..,,,

First we have to calculate the amount of solute required to prepare 1M NaOH standard solution

M=n/V(in L)

n=MxV

=1x1(M=1, V=1L)

n=1

Mass of NaOH /Molar mass of NaOH =1

Mass of NaOH=1 X Molar mass of NaOH

=1x40(molar mass of NaoH = 40 g/mol)

=40g

Hence we need 40g NaOH to prepare !M standard solution

Preparing a Standard solution video follows:-

https://youtu.be/ipyyrnjxkgy

DILUTION OF CONCENTRATED SOLUTIONS

Dilution law

M1V1=M2V2

During dilution of an acid, Add Acid to

Water(AAW),but not opposite?

Video link for How to dilute a concentrated

solutions follows:-

https://youtube/7AqfAqh5oSY

3. Calculate the mole fraction of ethylene glycol (C2H6O2) in a

solution containing 20% of C2H6O2 by mass.

Ans.

Solution will contain 20 g of ethylene glycol and 80 g of water.

Molar mass of C2H6O2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 g /mol

Moles of C2H6O2 = 20 g/62 g mol

= 0.322 mol

Moles of water = 80 g/18 g mol

= 4.444 mol

Mole fraction of C2H6O2=(moles of C2H6O2 /moles of

C2H6O2+moles of water)

=0.322/0.322+4.444

=0.068

Mole fraction of water=1-0.068 (mole fraction of C2H6O2 + mole

fraction of water=1)

= 0.932

4. Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in

75 g of benzene.

Ans. Molar mass of CH3COOH : 12 × 2 + 1 × 4 + 16 × 2 = 60 g /mol

Moles of CH3COOH = 2.5 g/60 g /mol = 0.0417 mol

Mass of benzene(solvent) in kg =75 g/1000 g/ kg

= 0.075kg

Molality(m) of CH3COOH = Moles of CH3COOH

Mass of benzene(solvent) in kg

= 0.0417 mol

0.075kg

= 0.556 mol/kg

5. CALCULATE THE MASS PERCENTAGE OF BENZENE (C6H6) AND CARBON

TETRACHLORIDE(CCl4) IF 22 g OF BENZENE IS DISSOLVED IN 122 g OF CARBON

TETRACHLORIDE.

Ans.

Mass percentage of benzene= Mass of benzene X100

Mass of solution

=22gX100

144g

=15.28%

Mass percentage of carbon tetrachloride=100-15.28

=84.72%

6. CALCULATE THE MOLE FRACTION OF BENZENE IN SOLUTION

CONTAINING 30% BY MASS IN CARBON TETRACHLORIDE.

Mole fraction of benzene= No. of moles of benzene

(No. of moles of benzene+No.of

moles of carbon tetrachloride)

No. of moles of benzene= mass of benzene/molar mass of

benzene

= 30g/78g/mol(molar mass of benzene

78g/mol)

=0.385

No. of moles of carbon tetrachloride= mass of carbon

tetrachloride/molar mass of

carbon tetrachloride

=70g/154g/mol

=0.455

Mole fraction of benzene=0.385/(0.385+0.455)

=0.458

Mole fraction of carbon tetrachloride=1-0.458

=0.542

7. CALCULATE THE MOLARITY OF EACH OF THE FOLLOWING SOLUTIONS: (A) 30 g OF

CO(NO3)2. 6H2O IN 4.3 L OF SOLUTION (B) 30 ML OF 0.5 M H2SO4 DILUTED TO 500 ML.

Ans.

(A)Molarity of solution= No. of moles of solute

Volume of solution in L

No. of moles of solute= Mass of Co(NO3)2. 6H2O

Molar mass of Co(NO3)2.6H2O

=30g/310.7g/mol

=0.0966

Volume of solution in L=4.3L

Molarity of solution=0.0966/4.3

=0.022M

(B)M1V1=M2V2

M1=0.5M,V1=30ML, V2=500ML, M2=?

M2=0.5X30/500

=15/500

=0.03M

8.CALCULATE THE MASS OF UREA (NH2CONH2) REQUIRED IN

MAKING 2.5 Kg OF 0.25 MOLAL AQUEOUS SOLUTION.

Molality(m)=No. of moles of solute(n)

Mass of the solvent(in Kg)

0.25=x/60 (x is mass of urea)

2.5

x=15g

Mass of solution=mass of solvent+mass of solute

=1000 + 15=1015g

1.015kg of solution contains 15g urea

2.5kg of solution contains = 2.5X15/1.015=37g

9.CALCULATE (A) MOLALITY (B) MOLARITY AND (C) MOLE FRACTION OF KI

IF THE DENSITY OF 20% (MASS/MASS) AQUEOUS KI IS 1.202 g/mL.

Ans. Here Mass of KI=20g,

Mass of solution=100g,

Mass of solvent=80g

(A)Molality(m)=0.12/0.080=1.5m

(B) Molarity(M)=0.12/0.0832=1.44M

(volume of solution=mass of solution/density)

=100/1.202=83.2ml=0.0832L)

(c) Mole fraction of KI= No.of moles of KI

Total no. of moles in solution

= 0.12/(0.12+4.44)

=0.0263

HOME WORK

2.1 Define the term solution. How many types of

solutions are formed? Write briefly about each type

with an example.

2.2 Give an example of a solid solution in which the

solute is a gas.

2.3 Define the following terms:

(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass

percentage.

2.4 Concentrated nitric acid used in laboratory work is

68% nitric acid by mass in aqueous solution. What

should be the molarity of such a sample of the acid if

the density of the solution is 1.504 g/ mL?

2.5 A solution of glucose in water is labelled as 10% w/w,

what would be the

molality and mole fraction of each component in the

solution? If the density of

solution is 1.2 g /Ml, then what shall be the molarity of the

solution?

2.6 How many mL of 0.1 M HCl are required to react

completely with 1 g mixture

of Na2CO3 and NaHCO3 containing equimolar amounts of

both?

2.7 A solution is obtained by mixing 300 g of 25% solution

and 400 g of 40%

solution by mass. Calculate the mass percentage of the

resulting solution.

2.8 An antifreeze solution is prepared from 222.6 g of

ethylene glycol (C2H6O2) and 200 g of water. Calculate

the molality of the solution. If the density of the

solution is 1.072 g /mL, then what shall be the

molarity of the solution?

2.9 A sample of drinking water was found to be

severely contaminated with

chloroform (CHCl3) supposed to be a carcinogen. The

level of contamination

was 15 ppm (by mass):

(i) express this in percent by mass

(ii) determine the molality of chloroform in the water

sample.

Thank you