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Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 8 April, Evening Shift — Question 3

HA(aq)⇌H+(aq)+A−(aq)\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) The freezing point depression of a 0.1 m aqueous solution of a monobasic

weak acid HA is 0.20∘C0.20^{\circ} \mathrm{C}. The dissociation constant for the acid is Given: Kf(H2O)=1.8 K kg mol−1\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1},

molality ≡\equiv molarity

  1. Option A:

    1.1×10−21.1 \times 10^{-2}

  2. Option B:

    1.38×10−31.38 \times 10^{-3}

    Correct
  3. Option C:

    1.90×10−31.90 \times 10^{-3}

  4. Option D:

    1.89×10−11.89 \times 10^{-1}

Answer: B

Step-by-step solution

ΔTf=iKfm\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m}

i=ΔTfKf⋅m\mathrm{i}=\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{f}} \cdot \mathrm{m}}

i=0.201.8×0.1=1.11i=\frac{0.20}{1.8 \times 0.1}=1.11

i=1.11\mathrm{i}=1.11

α=i−1n−1\alpha=\frac{\mathrm{i}-1}{\mathrm{n}-1} (for HA, n=2\mathrm{n}=2 )

α=1.11−11=0.11\alpha=\frac{1.11-1}{1}=0.11

Ka=cα21−α=0.1×(0.11)21−0.11=1.38×10−3K_{a}=\frac{c \alpha^{2}}{1-\alpha}=\frac{0.1 \times(0.11)^{2}}{1-0.11}=1.38 \times 10^{-3}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Abnormal Colligative Properties - van't Hoff Factor

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