When 9.45 g of ${\mathrm{ClCH}}_2\mathrm{COOH}$ is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ${\mathrm{ClCH}}_2\mathrm{COOH}$ is  $\mathrm{x}\times {10}^{-3}.$The value of x is ________. (Rounded off to the nearest integer).

$\left[{\mathrm{K}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.86\mathrm{K}\mathrm{kg}{\mathrm{mol}}^{-1}\right]$          

Moles of ClCH₂COOH:

$$\begin{gathered} \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}} \\ \mathrm{moles}=\frac{9.45 \mathrm{g}}{94.5 \mathrm{g}/\mathrm{mol}}=0.1 \mathrm{mol} \end{gathered}$$

$$\begin{gathered} {\mathrm{ClCH}}_2\mathrm{COOH}\rightleftharpoons {\mathrm{ClCH}}_2{\mathrm{COO}}^{-}+ {\mathrm{H}}^{+} \\ \mathrm{initial} 1 0 0 \\ \mathrm{At} \mathrm{eq}. \mathrm{C}\left(1-\mathrm{\alpha }\right) \mathrm{C\alpha } \mathrm{C\alpha } \\ \mathrm{i}=1+\mathrm{\alpha } \end{gathered}$$

$\begin{array}{l}{\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}\times {\mathrm{k}}_{\mathrm{f}}\times \mathrm{m}\\ 0.5=(1+\mathrm{\alpha })\times (1.86)\times \frac{9.45\times 1000}{94.5\times 500}\\ 0.5=(1+\mathrm{\alpha })\times 1.86\times \frac{1}{5}\\ (1+\mathrm{\alpha })=\frac{0.5}{1.86}\\ \mathrm{\alpha }=0.34\end{array}$

Dissociation constant (K_a) for ClCH₂COOH:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{C\alpha }.\mathrm{C\alpha }}{\mathrm{C}(1-\mathrm{\alpha })} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{0.2\times {\left(0.34\right)}^2}{1-0.34} \\ {\mathrm{K}}_{\mathrm{a}}=\underset{\mathrm{x}}{\underbrace{35}}\times {10}^{-3} \end{gathered}$$