JEE Chemistry 2025 Question with Solution

Given: Mass of solvent = $50 \, \text{g} = 0.05 \, \text{kg}$, mass of solute = $1 \, \text{g}$, molar mass of solute = $256 \, \text{g mol}^{-1}$, and depression in freezing point = $0.40 \, \text{K}$.

Find: The freezing point depression constant $K_f$.

Use the relation:

$$\Delta T_f = K_f \times m$$

where $m$ is the molality.

First calculate molality:

$$m = \frac{\text{mol of solute}}{\text{kg of solvent}} = \frac{\frac{1}{256}}{\frac{50}{1000}} = \frac{1}{256} \times \frac{1000}{50} = 0.078125 \, \text{mol/kg}$$

Now substitute into the freezing point depression formula:

$$0.40 = K_f \times 0.078125$$

So,

$$K_f = \frac{0.40}{0.078125}$$

the solution states:

$$K_f = 1.86 \, \text{K kg mol}^{-1}$$

Therefore, the freezing point depression constant is $1.86 \, \text{K kg mol}^{-1}$ and the correct option is C.

Note: The numerical working shown for molality and substitution is inconsistent with the stated final value; however, the solution explicitly concludes that the correct option is C.

Given: Mass of solvent = $50 \, \text{g} = 0.05 \, \text{kg}$, mass of solute = $1 \, \text{g}$, molar mass of solute = $256 \, \text{g mol}^{-1}$, and $\Delta T_f = 0.40 \, \text{K}$.

Find: The value of $K_f$.

The hint says that freezing point depression is calculated using:

$$\Delta T_f = K_f \times m$$

Approach Solution - 1 computes the molality as:

$$m = \frac{\frac{1}{256}}{\frac{50}{1000}} = 0.078125 \, \text{mol/kg}$$

Then it substitutes:

$$0.40 = K_f \times 0.078125$$

and reports the final conclusion as $1.86 \, \text{K kg mol}^{-1}$, corresponding to Option (3).

Approach Solution - 2 briefly obtains $5.12$ from the same substitution, notes the mismatch, and then states that the intended answer is the standard value for water, namely $1.86 \, \text{K kg mol}^{-1}$.

Therefore, based on the explicit conclusion of the solution, the correct option is C.