NEETprep Answer:

Molar mass of ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHClCOOH}=15+14+13+35.5+12+16+16+1=122.5 \mathrm{g} {\mathrm{mol}}^{-1}$

$\therefore$No. of moles present in 10 g of ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHClCOOH}=\frac{10 \mathrm{g}}{122.5 \mathrm{g} {\mathrm{mol}}^{-1}}$${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHCICOOH}=\frac{10 \mathrm{g}}{122.5 \mathrm{g} {\mathrm{mol}}^{-1}}$

$=0.0816 \mathrm{mol}$

It is given that 10 g of is added to 250 g of water.

Molality of the solution,

Let α be the degree of dissociation of CH₃CH₂CHClCOOH.

CH₃CH₂CHClCOOH undergoes dissociation according to the following equation: 

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHCICOOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CHCICOOH}}^{-}+{\mathrm{H}}^{+}$
$\mathrm{Initial} \mathrm{conc}. \mathrm{C} \mathrm{mol} {\mathrm{L}}^{-1 } 0 0$
$\mathrm{At} \mathrm{equlibrium} \mathrm{C}(1-\mathrm{a}) \mathrm{Ca} \mathrm{Ca}$
$\therefore {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{Ca}.\mathrm{Ca}}{\mathrm{C}(1-\mathrm{a})}$
$=\frac{{\mathrm{Ca}}^2}{1-\mathrm{a}}$

Since α is very small with respect to 1, 1 − α ≈ 1

Now, ${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{Ca}}^2}{1}$

$\Rightarrow {\mathrm{K}}_{\mathrm{a}}={\mathrm{C\alpha }}^2$
$\Rightarrow \mathrm{a}=\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}$
$=\frac{1.4\times {10}^{-3}}{0.3264} (\therefore {\mathrm{K}}_{\mathrm{a}}=1.4\times {10}^{-3})$
$=0.0655$

Again,

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHClCOOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CHClCOOH}}^{-}+{\mathrm{H}}^{+}$
$\mathrm{Initial} \mathrm{conc}. 1 0 0$
$\mathrm{At} \mathrm{equilibrium} 1-\mathrm{\alpha } \alpha \alpha$

Total moles of equilibrium = 1 − α + α + α = 1 + α

$\therefore \mathrm{i}=\frac{1+\mathrm{\alpha }}{1}$
$=1+\mathrm{\alpha }$
$=1+0.0655$
$=1.0655$

Hence, the depression in the freezing point of water is given as:

$∆{\mathrm{T}}_{\mathrm{f}}=\mathrm{i}.{\mathrm{K}}_{\mathrm{f}}\mathrm{m}$
$=1.0665\times 1.86 \mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}\times 0.3264 \mathrm{mol} {\mathrm{kg}}^{-1}$
$=0.65 \mathrm{K}$