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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 5 April · Evening Shift · Chemistry · Solutions

Considering acetic acid dissociates in water, its dissociation...

Considering acetic acid dissociates in water, its dissociation constant is (6.25 \times 10^{-5}). If (5 \mathrm{ mL}) of acetic acid is dissolved in 1 litre water, the solution will freeze at (-x \times 10^{-2}{ }^{\circ} \mathrm{C}), provided pure water freezes at (0^{\circ} \mathrm{C}). x=x= ______ . (Nearest integer) Given : (Kf)water =1.86Kkgmol−1.  density of  acetic acid is 1.2gmol−1.  molar mass of water =18gmol−1.  molar mass of acetic acid=60 gmol−1.  density of water =1gcm−3\begin{aligned} & \left(\mathrm{K}_f\right)_{\text{water }}=1.86 \mathrm{ K} \mathrm{ kg} \mathrm{ mol}^{-1} \text{. } \\ & \text{ density\text{ of } acetic acid is } 1.2 \mathrm{ g} \mathrm{ mol}^{-1} \text{. } \\ & \text{ molar mass of water }=18 \mathrm{ g} \mathrm{ mol}^{-1} \text{. } \\ & \text{ molar mass of acetic acid=60 } \mathrm{g} \mathrm{mol}^{-1} \text{. } \\ & \text{ density of water }=1 \mathrm{ g} \mathrm{ cm}^{-3} \end{aligned}) Acetic acid dissociates as (\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus})$

Correct answer

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Step-by-step solution

 Mass of  CH3COOH=V×d=5ml×1.2g/ml=6gmnCH3COOH=660=0.1molmCH3COOH≈MCH3COOH=0.11=0.1MCH3COOH⇌CH3COO−+H+\begin{aligned} & \text{ Mass\text{ of } } \mathrm{CH}_3 \mathrm{COOH}=\mathrm{V} \times \mathrm{d} \\ & =5 \mathrm{ml} \times 1.2 \mathrm{ g} / \mathrm{ml} \\ & =6 \mathrm{gm} \\ & \mathrm{n}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{6}{60}=0.1 \mathrm{ mol} \\ & \mathrm{ m}_{\mathrm{CH}_3 \mathrm{COOH}} \approx \mathrm{M}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{0.1}{1}=0.1 \mathrm{M} \\ & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\end{aligned} C C−CαCαCαKa=Cα21−α1−α≈1⇒Ka=Cα2α=KaC=6.25×10−50.1=25×10−3\begin{aligned} & \mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\ & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 1-\alpha \approx 1 \Rightarrow \mathrm{K}_{\mathrm{a}}=\mathrm{C} \alpha^2 \\ & \alpha=\sqrt{\frac{\mathrm{Ka}}{\mathrm{C}}}=\sqrt{\frac{6.25 \times 10^{-5}}{0.1}}=25 \times 10^{-3}\end{aligned} V.f. (i)=1+α(n−1)=1+α(2−1)=1+α=1+25×10−3=1.025ΔTf=iKfm=(1.025)(1.86)(0.1)=0.19=19×10−2\begin{aligned} & \text{ V.f. }(\mathrm{i})=1+\alpha(\mathrm{n}-1)=1+\alpha(2-1)=1+\alpha \\ & =1+25 \times 10^{-3}=1.025 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\ & =(1.025)(1.86)(0.1) \\ & =0.19 \\ & =19 \times 10^{-2}\end{aligned}$
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