JEE Main 2024 — Chemistry Physical Chemistry

$\begin{aligned} & \text { Mass of } \mathrm{CH}_3 \mathrm{COOH}=\mathrm{V} \times \mathrm{d} \\ & =5 \mathrm{ml} \times 1.2 \mathrm{~g} / \mathrm{ml} \\ & =6 \mathrm{gm} \\ & \mathrm{n}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{6}{60}=0.1 \mathrm{~mol} \\ & \mathrm{~m}_{\mathrm{CH}_3 \mathrm{COOH}} \approx \mathrm{M}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{0.1}{1}=0.1 \mathrm{M} \\ & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\end{aligned}$​ Mass of CH3​COOH=V×d=5ml×1.2 g/ml=6gmnCH3​COOH​=606​=0.1 mol mCH3​COOH​≈MCH3​COOH​=10.1​=0.1MCH3​COOH⇌CH3​COO−+H+​ C $\begin{aligned} & \mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\ & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 1-\alpha \approx 1 \Rightarrow \mathrm{K}_{\mathrm{a}}=\mathrm{C} \alpha^2 \\ & \alpha=\sqrt{\frac{\mathrm{Ka}}{\mathrm{C}}}=\sqrt{\frac{6.25 \times 10^{-5}}{0.1}}=25 \times 10^{-3}\end{aligned}$​C−CαCαCαKa​=1−αCα2​1−α≈1⇒Ka​=Cα2α=CKa​​=0.16.25×10−5​​=25×10−3​ $\begin{aligned} & \text { V.f. }(\mathrm{i})=1+\alpha(\mathrm{n}-1)=1+\alpha(2-1)=1+\alpha \\ & =1+25 \times 10^{-3}=1.025 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\ & =(1.025)(1.86)(0.1) \\ & =0.19 \\ & =19 \times 10^{-2}\end{aligned}$​ V.f. (i)=1+α(n−1)=1+α(2−1)=1+α=1+25×10−3=1.025ΔTf​=iKf​m=(1.025)(1.86)(0.1)=0.19=19×10−2​