$12.25 \mathrm{~g}$ of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHClCOOH}$ is added to $250 \mathrm{~g}$ of…

1. 0.789
2. 0.394
3. 1.183
4. 0.592

Solution

Step I Calculation of degree of dissociation Molar concentration of solution $ \begin{aligned} & =\frac{\text { Mass of solute }}{\text { Molar mass of solute } \times \text { Mass of solvent }} \times 1000 \\ & =\frac{12.25 \mathrm{~g} \times 1000}{122.5 \mathrm{~g} \mathrm{~mol}^{-1} \times 250 \mathrm{~g}} \\ & \begin{aligned} & \because \text { Molecular weight of solute }\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHClCOOH}\right) \\ = & =122.5 \mathrm{~g} / \mathrm{mol}] \end{aligned} \\ & =0.40 \mathrm{~m} \end{aligned} $ If $\alpha$ is the degree of dissociation of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHClCOOH}$, then
$ \begin{aligned} & \therefore K_a=\frac{C \alpha^2}{(1-\alpha)} \approx C \alpha^2 \\ & \Rightarrow \alpha=\sqrt{\frac{K_a}{C}} \\ & \Rightarrow \alpha=\sqrt{\frac{1.44 \times 10^{-3}}{0.4}}=\sqrt{36 \times 10^{-4}}=0.06 \end{aligned} $ Step II Calculation of van't-Hoff factor $ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHClCOOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHClCOO}^{-}+\mathrm{H}^{+} $ Initial moles 1 $ 00 $ Conc. equilibrium at $1-\alpha \quad \alpha \quad \alpha$ Total number of moles after dissociation $ =1-\alpha+\alpha+\alpha=1+\alpha $ $\therefore$ van't-Hoff factor $ \begin{aligned} (i) & =\frac{\text { Total number of moles after dissociation }}{\text { Number of moles before dissociation }} \\ & i=\frac{(1+\alpha)}{1}=1+\alpha \\ \therefore i & =1+0.06=1.06 \end{aligned} $ Step III Calculation of depression in freezing point $(\Delta)$. $ \begin{aligned} \Delta T_f & =i K_f m=(1.06) \times 0.40 \times 1.86 \\ \therefore \quad \Delta T_f & =0.789 \mathrm{~K} \end{aligned} $

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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