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Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 5 April, Shift 2 — Question 87

Considering acetic acid dissociates in water, its dissociation constant is 6.25×10−56.25 \times 10^{-5}. If 5 mL of acetic acid is dissolved

in 1 litre water, the solution will freeze at −x×10−2∘C-\mathrm{x} \times 10^{-2}{ }^{\circ} \mathrm{C}, provided pure water freezes at 0∘C0^{\circ} \mathrm{C}. x=\mathrm{x}= (Nearest integer)

Given : (Kf)water =1.86 K kg mol−1\left(\mathrm{K}_{\mathrm{f}}\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}. density of acetic acid is 1.2 g mol−11.2 \mathrm{~g} \mathrm{~mol}^{-1}

molar mass of water =18 g mol−1=18 \mathrm{~g} \mathrm{~mol}^{-1}.

molar mass of acetic acid =60 g mol−1=60 \mathrm{~g} \mathrm{~mol}^{-1}. density of water =1 g cm−3=1 \mathrm{~g} \mathrm{~cm}^{-3}

Acetic acid dissociates as CH3COOH⇌CH3COO⊕+H⊕\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{\oplus}+\mathrm{H}^{\oplus}

Answer: 19

Numerical answer — enter this value.

Step-by-step solution

Mass of CH3COOH=V×d\mathrm{CH}_{3} \mathrm{COOH}=\mathrm{V} \times \mathrm{d}

=5ml×1.2 g/ml=5 \mathrm{ml} \times 1.2 \mathrm{~g} / \mathrm{ml}

=6gm=6 \mathrm{gm}

nCH3COOH=660=0.1 mol\mathrm{n}_{\mathrm{CH}_{3} \mathrm{COOH}}=\frac{6}{60}=0.1 \mathrm{~mol}

mCH3COOH≈MCH3COOH=0.11=0.1M\mathrm{m}_{\mathrm{CH}_{3} \mathrm{COOH}} \approx \mathrm{M}_{\mathrm{CH}_{3} \mathrm{COOH}}=\frac{0.1}{1}=0.1 \mathrm{M}

CH3COOH⇌CH3COO−+H+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}

figure

Ka=Cα21−α\mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}

1−α≈1⇒ Ka=Cα21-\alpha \approx 1 \Rightarrow \mathrm{~K}_{\mathrm{a}}=\mathrm{C} \alpha^{2}

α=KaC=6.25×10−50.1=25×10−3\alpha=\sqrt{\frac{\mathrm{Ka}}{\mathrm{C}}}=\sqrt{\frac{6.25 \times 10^{-5}}{0.1}}=25 \times 10^{-3} V.f. (i)=1+α(n−1)=1+α(2−1)=1+α(i)=1+\alpha(n-1)=1+\alpha(2-1)=1+\alpha

=1+25×10−3=1.025=1+25 \times 10^{-3}=1.025

ΔTf=iKm\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK} \mathrm{m}

=(1.025)(1.86)(0.1)=(1.025)(1.86)(0.1)

=0.19=0.19

=19×10−2=19 \times 10^{-2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 2
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Abnormal Colligative Properties - van't Hoff Factor