Q25 / 90 · JEE Main 2024 (05 Apr Shift 2) · Chemistry
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 \(0^{\circ} \mathrm{C}\).
\(x=\) ______ . (Nearest integer)
Given :
\(\begin{aligned}
& \left(\mathrm{K}_f\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \text {. } \\
& \text { density of acetic acid is } 1.2 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\
& \text { molar mass of water }=18 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\
& \text { molar mass of acetic acid=60 } \mathrm{g} \mathrm{mol}^{-1} \text {. } \\
& \text { density of water }=1 \mathrm{~g} \mathrm{~cm}^{-3}
\end{aligned}\)
Acetic acid dissociates as \(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus}\)
Correct answer: 19
Solution
$\begin{aligned} & \text { Mass of } \mathrm{CH}_3 \mathrm{COOH}=\mathrm{V} \times \mathrm{d} \\ & =5 \mathrm{ml} \times 1.2 \mathrm{~g} / \mathrm{ml} \\ & =6 \mathrm{gm} \\ & \mathrm{n}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{6}{60}=0.1 \mathrm{~mol} \\ & \mathrm{~m}_{\mathrm{CH}_3 \mathrm{COOH}} \approx \mathrm{M}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{0.1}{1}=0.1 \mathrm{M} \\ & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\end{aligned}$
C
$\begin{aligned} & \mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\ & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 1-\alpha \approx 1 \Rightarrow \mathrm{K}_{\mathrm{a}}=\mathrm{C} \alpha^2 \\ & \alpha=\sqrt{\frac{\mathrm{Ka}}{\mathrm{C}}}=\sqrt{\frac{6.25 \times 10^{-5}}{0.1}}=25 \times 10^{-3}\end{aligned}$
$\begin{aligned} & \text { V.f. }(\mathrm{i})=1+\alpha(\mathrm{n}-1)=1+\alpha(2-1)=1+\alpha \\ & =1+25 \times 10^{-3}=1.025 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\ & =(1.025)(1.86)(0.1) \\ & =0.19 \\ & =19 \times 10^{-2}\end{aligned}$
Hence, the correct answer is 19.