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Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 4 April, Evening Shift — Question 6

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II:

Statement I: Molal depression constant Kf\mathrm{K}_{\mathrm{f}} is given by M1RTfΔSfus\frac{M_{1} R T_{f}}{\Delta S_{f u s}}, where symbols have their usual meaning.

Statement II: Kf\mathrm{K}_{\mathrm{f}} for benzene is less than the Kf\mathrm{K}_{\mathrm{f}} for water.

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

  1. Option A:

    Statement I is correct but Statement II is incorrect

    Correct
  2. Option B:

    Both Statement I and Statement II are correct

  3. Option C:

    Statement I is incorrect but Statement II is correct

  4. Option D:

    Both Statement I and Statement II are incorrect

Answer: A

Step-by-step solution

∵ΔSfus =ΔHfus T\because \Delta \mathrm{S}_{\text {fus }}=\frac{\Delta \mathrm{H}_{\text {fus }}}{\mathrm{T}}

So, Kf:MRTf2ΔHfus=MRTfΔSfus K_{f}: \frac{M R T_{f}^{2}}{\Delta H_{f u s}}=\frac{M R T_{f}}{\Delta S_{\text {fus }}}

Kf(H2O)=1.86 K kg mol−1\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

Kf(\mathrm{K}_{\mathrm{f}}( benzene )=5.12 K mol−1)=5.12 \mathrm{~K} \mathrm{~mol}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Solid in Liquid Solutions (Colligative Properties)
Given below are two statements : one is labelled as Statement I and… | JEE Main 2025 PYQ with Solution · DhiX AI