JEE Chemistry 2023 Question with Solution

Given: Mass of solute = $2 \, \text{g}$, mass of water = $20 \, \text{g} = 0.02 \, \text{kg}$, boiling point of solution = $373.52 \, \text{K}$, normal boiling point of water = $373 \, \text{K}$, and $K_b = 0.52 \, \text{K kg mol}^{-1}$.

Find: Molecular mass of the non-volatile solute.

First, calculate the elevation in boiling point:

$$\Delta T_b = T_b - T_b^\circ = 373.52 - 373 = 0.52 \, \text{K}$$

Using the boiling point elevation relation:

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

Substitute molality:

$$m = \frac{\text{mass of solute (g)}}{\text{Molar Mass (g mol}^{-1}) \times \text{mass of solvent (kg)}}$$

So,

$$0.52 = 0.52 \cdot \frac{2}{\text{Molar Mass} \times 20 \times 10^{-3}}$$

Simplifying,

$$1 = \frac{2}{\text{Molar Mass} \times 0.02}$$

Therefore,

$$\text{Molar Mass} = \frac{2}{0.02} = 100 \, \text{g mol}^{-1}$$

Therefore, the molecular mass of the solute is $100 \, \text{g mol}^{-1}$.

Given: $\Delta T_b = 0.52 \, \text{K}$ and $K_b = 0.52 \, \text{K kg mol}^{-1}$.

Find: Molar mass of the solute.

From

$$\Delta T_b = K_b m$$

we get

$$m = \frac{\Delta T_b}{K_b} = \frac{0.52}{0.52} = 1 \, \text{mol kg}^{-1}$$

Now molality is also

$$m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{2/M}{0.02}$$

where $M$ is the molar mass.

Thus,

$$1 = \frac{2/M}{0.02}$$ $$0.02 = \frac{2}{M}$$ $$M = \frac{2}{0.02} = 100$$

Hence, the required molecular mass is $100 \, \text{g mol}^{-1}$.