JEE Main 2025 — Solutions Question with Solution

To find the freezing point depression constant ($K_f$) of the solvent, we use the formula for freezing point depression:

$\Delta T_f=K_f\cdot m$

Given:

The decrease in freezing point $\Delta T_f$ is 0.40 K.

The mass of the solute is 1 g and its molar mass is 256 g/mol.

The mass of the solvent is 50 g (or 0.050 kg).

First, calculate the molality ($m$):

Molality is defined as the moles of solute per kilogram of solvent.

Calculate moles of solute:

$\text{Moles of solute}=\frac{1\text{g}}{256\text{g/mol}}=\frac{1}{256}\text{mol}$

Calculate molality ($m$):

$m=\frac{\frac{1}{256}\text{mol}}{0.050\text{kg}}=\frac{1}{256\times 0.050}\text{mol/kg}$

Now, substitute into the formula to find $K_f$:

$0.4=K_f\cdot \frac{1}{256\times 0.050}$

Solving for $K_f$:

$K_f=0.4\cdot 256\times 0.050=5.12\text{K kg/mol}$

Thus, the freezing point depression constant of the solvent is $5.12\text{K kg/mol}$.