Ncert Exercises & Solutions

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. The molar mass of the solute is -

1. 53.50 g mol^-1

2. 43.5 kg mol^-1

3. 22.6 kg mol-1

4. 41.35 g mol^-1

Hint: Use relative lowering of vapour pressure formula

Step 1:

Here,

Vapour pressure of the solution at normal boiling point (p₁) = 1.004 bar

Vapour pressure of pure water at normal boiling point $(p_{1}^{0}) = 1.013 bar$

2% non-volatile solute means at 100 g solution 2g solute is present in the solution.

Mass of solute, (w₂) = 2 g

Mass of solvent (water), (w₁) = 98 g

The molar mass of solvent (water), (M₁) = 18 g mol⁻¹

Step 2:

According to Raoult’s law,

$\frac{P_{1}^{o}-P_{1}}{P_{1}^{o}}=X_{solute}$

$\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$
$\Rightarrow \frac{1.013 - 1.004}{1.013} = \frac{2 \times 18}{M_{2} \times 98}$
$\Rightarrow \frac{0.009}{1.013} = \frac{2 \times 18}{M_{2} \times 98}$
$\Rightarrow M_{2} = \frac{1.013 \times 2 \times 18}{0.009 \times 98}$
$= 41.35 g {mol}^{- 1}$

Hence, the molar mass of the solute is 41.35 g mol^-1

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions