JEE Main 2025 Apr 4 Shift 2, Chemistry Q7: Colligative Properties

JEE Main2025Apr 4, Shift 2Chemistry

Given below are two statements :

Statement (I) : Molal depression constant $K_{f}$ is given by $\frac{M _{1} R T _{f}}{Δ S _{f u s}}$ , where symbols have their usual meaning.

Statement (II) : $K_{f}$ for benzene is less than the $K_{f}$ for water.

In the light of the above statements, choose the most appropriate answer from the options given below :

A
Statement I is incorrect but Statement II is correct
B
Both Statement I and Statement II are incorrect
C
Both Statement I and Statement II are correct
D
Statement I is correct but Statement II is incorrect

Solution

Statement I. Starting from $K_{f} = \frac{M _{1} R T _{f}^{2}}{Δ H _{f u s}}$ and using $Δ S_{f u s} = \frac{Δ H _{f u s}}{T _{f}}$ , we get $K_{f} = \frac{M _{1} R T _{f}}{Δ S _{f u s}}$ . The expression in Statement I is therefore correct.

Statement II. The standard tabulated values are $K_{f} (benzene) = 5.12 K kg mol^{- 1}$ and $K_{f} (water) = 1.86 K kg mol^{- 1}$ . So $K_{f}$ for benzene is actually greater than that for water; Statement II is incorrect.

$∴$ Statement I correct, Statement II incorrect.

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JEE Main 2025 Apr 4 Shift 2, Chemistry Q7: Colligative Properties

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