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Chemistry · Ionic Equilibrium

JEE Main 2026 — 2 April, Morning Shift — Question 44

19.5 g of fluoro acetic acid (molar mass =78 g mol−1=78 \mathrm{~g} \mathrm{~mol}^{-1} ) is dissolved in 500 g of water at 298 K . The depression in the freezing point of water was 1∘C1^{\circ} \mathrm{C}. What is Ka\mathrm{K}_{\mathrm{a}} of fluoro acetic acid ? (For water, Kf=1.86 K kg mol−1\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} ). Assume molarity and molality to have same values

  1. Option A:

    10−610^{-6}

  2. Option B:

    4×10−44 \times 10^{-4}

  3. Option C:

    3×10−53 \times 10^{-5}

  4. Option D:

    3×10−33 \times 10^{-3}

    Correct

Answer: D

Step-by-step solution

ΔTf=iKrm\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{r}} \mathrm{m} no. of moles =19.578=14=\frac{19.5}{78}=\frac{1}{4} mole 1=i×1.86×1/41/21=\mathrm{i} \times 1.86 \times \frac{1 / 4}{1 / 2} i=21.86\mathrm{i}=\frac{2}{1.86} i=1+(n−1)α\mathrm{i}=1+(\mathrm{n}-1) \alpha i=1+α\mathrm{i}=1+\alpha α=21.86−1=0.075\alpha=\frac{2}{1.86}-1=0.075 Ka=C21−α\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C}^{2}}{1-\alpha} =12×(0.075)2(1−0.075)=\frac{1}{2} \times \frac{(0.075)^{2}}{(1-0.075)} =3×10−3=3 \times 10^{-3}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Solutions containing one Acid or Base

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