• 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 cm³ mixture of CO, CH₄ and He gases and exploded it by an electric discharge at room temperature with excess of oxygen. The volume initially contracted to 13.0 cm³. Then it further contracted to 14.0 cm³ when the residual gas is treated with KOH solution. Find out the consumption of gaseous mixture in terms of volume percentage.

Solution:

Take V₁= Partial volumes of CO 

V₂= Partial volumes of CH₄

 V₃ = Partial volumes of He

V₁+V₂+V₃ = 20.0 cm³

The equation for combustion is as follows:

CO + ½ O₂ –> CO₂

CH₄ + 2O₂ –> CO₂ + 2H₂O

V₁ volume of CO = ½ V₁ volume of O₂ = V₁ volume of CO₂

V₂ volume of CH₄ = 2V₂ volume O₂ = V₂ volume of CO₂

V1/2 + 2V₂ = 13.0 cm³

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

V₁+ V₂ = 14.0 cm³

V₁ = 10 cm³

V₂ = 4.0 cm³

V₃ = 6.0 cm³

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

Volume % of CH₄ = 4/20 X 100 = 20

Volume % of He = 100 – (50 + 20) = 30

2.1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. (NCERT Book)
Ans: Mass of solution = Mass of C₆H₆ + Mass of CCl₄
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl₄ = 122/144 x 100 = 84.72 %

2.28. Calculate the mass percentage of aspirin (C₉H₈O₄ in acetonitrile (CH₃CN) when 6.5g of CHO is dissolved in 450 g of CH3CN. (NCERT Book)

Q.