JEE Chemistry 2026 Question with Solution

Given:

• Mass of solute X = $1.5 \, \text{g}$
• Mass of solvent Y = $150 \, \text{g} = 0.150 \, \text{kg}$
• Molar mass of solvent Y = $300 \, \text{g mol}^{-1}$
• Elevation in boiling point = $0.5 \, \text{K}$
• $K_b = 5.0 \, \text{K kg mol}^{-1}$**

Find: The relative lowering in vapour pressure of solvent Y.

Use the colligative property relation:

$$\Delta T_b = K_b \cdot m$$

So,

$$m = \frac{0.5}{5.0} = 0.1 \, \text{mol kg}^{-1}$$

Molality is moles of solute per kilogram of solvent, hence moles of solute are:

$$n_{\text{solute}} = m \times W_{\text{solvent(kg)}} = 0.1 \times 0.150 = 0.015 \, \text{mol}$$

Now calculate moles of solvent:

$$n_{\text{solvent}} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{150}{300} = 0.5 \, \text{mol}$$

For a dilute solution, relative lowering in vapour pressure is approximately equal to mole fraction of solute:

$$RLVP = \frac{P^{\circ} - P}{P^{\circ}} = X_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$$

Substituting values:

$$RLVP \approx \frac{0.015}{0.5} = 0.03$$

Therefore,

$$RLVP = 3 \times 10^{-2}$$

The required nearest integer is 3.**

Given: $\Delta T_b = 0.5 \, \text{K}$ and $K_b = 5.0 \, \text{K kg mol}^{-1}$.

Find: Relative lowering in vapour pressure.

First find molality:

$$m = \frac{\Delta T_b}{K_b} = \frac{0.5}{5.0} = 0.1 \, \text{mol kg}^{-1}$$

For dilute solutions, use the direct relation:

$$RLVP = \frac{m \cdot M_{\text{solvent}}}{1000}$$

where $M_{\text{solvent}} = 300 \, \text{g mol}^{-1}$.

So,

$$RLVP = \frac{0.1 \times 300}{1000} = 0.03 = 3 \times 10^{-2}$$

This shortcut works because for dilute solutions, mole fraction of solute can be written directly in terms of molality and molar mass of the solvent.

Hence, the required integer is 3.