Given below are two statements : Statement (I) : Molal depression constant $\mathrm{K}_{\mathrm{f}}$ is given by $\frac{\mathrm{M}_{1} \mathrm{RT}_{\mathrm{f}}}{\Delta \mathrm{S}_{\text {fus }}}$, where symbols have their usual meaning. Statement (II) : $\mathrm{K}_{\mathrm{f}}$ for benzene is less than the $\mathrm{K}_{\mathrm{f}}$ for water. In the light of the above statements, choose the most appropriate answer from the options given below :

(D) Statement I is correct but Statement II is incorrect

(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

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Solutions

Given below are two statements :

Statement (I) : Molal depression constant

K_{f}

is given by

\frac{M _{1} RT _{f}}{Δ S _{fus}}

, 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✓ Correct

Answer: (D)

Solution

The molal freezing-point depression constant is

K_{f} = \frac{M _{1} R T _{f}^{2}}{Δ H _{fus}}

, and since the entropy of fusion is

Δ S_{fus} = \frac{Δ H _{fus}}{T _{f}}

, substitution gives

K_{f} = \frac{M _{1} R T _{f}}{Δ S _{fus}}

, so the first expression is correct. Numerically

K_{f}

for benzene is

5.12 K kg mo l^{- 1}

against

1.86 K kg mo l^{- 1}

for water, so benzene has the larger constant and the claim that it is smaller than water's is false. The first statement is correct while the second is incorrect.

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Solutions

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