- Concentration of a solution is the measure of the composition of a solution.
- A solution with relatively very large quantity of solute is called concentrated solution.
- A solution with relatively very small quantity of solute is called a dilute solution.

**Problem:**

Ami took a 20 cm^{3} mixture of CO, CH_{4} and He gases and exploded it by an electric discharge at room temperature with excess of oxygen. The volume initially contracted to 13.0 cm^{3}. Then it further contracted to 14.0 cm^{3} when the residual gas is treated with KOH solution. Find out the consumption of gaseous mixture in terms of volume percentage.

**Solution:**

Take V_{1}= Partial volumes of CO

V_{2}= Partial volumes of CH_{4}

V_{3} = Partial volumes of He

V_{1}+V_{2}+V_{3} = 20.0 **cm ^{3}**

The equation for combustion is as follows:

CO + ½ O_{2} –> CO_{2}

CH_{4} + 2O_{2} –> CO_{2} + 2H_{2}O

V_{1} volume of CO = ½ V_{1} volume of O_{2} = V_{1} volume of CO_{2}

V_{2} volume of CH_{4} = 2V_{2} volume O_{2} = V_{2} volume of CO_{2}

V1/2 + 2V_{2} = 13.0 **cm ^{3}**

Treating of residual gases with KOH solution would cause the absorption of CO2

V_{1}+ V_{2} = 14.0 **cm ^{3}**

V_{1} = 10** cm ^{3}**

V_{2} = 4.0 **cm ^{3}**

V_{3} = 6.0 **cm ^{3}**

Volume % of CO = 10/20 x 100 = 50

Volume % of CH_{4} = 4/20 X 100 = 20

Volume % of He = 100 – (50 + 20) = 30

**2.1. Calculate the mass percentage of benzene (C _{6}H_{6}) and carbon tetrachloride (CCl_{4}) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. (NCERT Book)**

**Ans:**Mass of solution = Mass of C

_{6}H

_{6}+ Mass of CCl

_{4}

= 22 g+122 g= 144 g

Mass % of benzene = 22/144 x 100 =15.28 %

Mass % of CCl

_{4}= 122/144 x 100 = 84.72 %

**2.28. Calculate the mass percentage of aspirin (C _{9}H_{8}O_{4} in acetonitrile (CH_{3}CN) when 6.5g of CHO is dissolved in 450 g of CH3CN. (NCERT Book)**

Q.