A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y'(molar mass=300 g $$mol^{-1}$$ ) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is __________ $$\times 10^{-2}$$. (nearest integer)
[Given : $$K_{b}$$ of the solvent =5.0 K kg $$mol^{-1}$$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Elevation of boiling point: $$\Delta T_b = K_b \cdot m$$

$$0.5 = 5.0 \times \frac{w/M}{0.150}$$ where w = 1.5 g and M is molar mass of X.

$$\frac{w}{M} = \frac{0.5 \times 0.150}{5.0} = 0.015$$ mol

$$M = \frac{1.5}{0.015} = 100$$ g/mol

Relative lowering of vapor pressure: $$\frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$$ (dilute)

$$n_{\text{solute}} = 0.015$$ mol, $$n_{\text{solvent}} = \frac{150}{300} = 0.5$$ mol

$$\frac{\Delta P}{P^0} = \frac{0.015}{0.5} = 0.03 = 3 \times 10^{-2}$$

The answer is 3 × 10⁻².