When $9.45 \mathrm{~g}$ of $\mathrm{ClCH}_{2} \mathrm{COOH}$ is added to 500 $\mathrm{mL}$ of water, its freezing point drops by $0.5^{\circ} \mathrm{C}$. The dissociation constant of $\mathrm{Cl} \mathrm{CH}_{2} \mathrm{COOH}$ is $x \times 10^{-3}$. The value of $x$ is____________
(Rounded off to the nearest integer)
$\left[\mathrm{K}_{\mathrm{f}\left(\mathrm{H}_{2} \mathrm{O}\right)}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$
$\mathrm{ClCH}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{ClCH}_{2} \mathrm{COO}^{\odot}+\mathrm{H}^{+}$
$\mathrm{i}=1+(2-1) \alpha$
$\mathrm{i}=1+\alpha$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \mathrm{m}$
$0.5=(1+\alpha)(1.86)\left(\frac{\left(\frac{9.45}{94.5}\right)}{\left(\frac{500}{1000}\right)}\right)$
$\frac{5}{3.72}=1+\alpha \Rightarrow \alpha=\frac{1.28}{3.72}$
$\alpha=\frac{32}{93}$
$\begin{array}{lrr}\mathrm{ClCH}_{2} \mathrm{COOH} & \rightleftharpoons \mathrm{ClCH}_{2} & \mathrm{COO}^{\odot}+\mathrm{H}^{+} \\ \mathrm{C}-\mathrm{C \alpha} & \mathrm{C} \alpha & \mathrm{C} \alpha\end{array}$
$\mathrm{K}_{\mathrm{a}}=\frac{(\mathrm{C} \alpha)^{2}}{\mathrm{C}-\mathrm{C} \alpha}=\frac{\mathrm{C} \alpha^{2}}{1-\alpha}$
$\mathrm{C}=\frac{0.1}{500 / 1000}=0.2$
$\mathrm{K}_{\mathrm{a}}=\frac{0.2(32 / 93)^{2}}{(1-32 / 93)}=\frac{0.2 \times(32)^{2}}{93 \times 61}$
$=0.036$
$\mathrm{~K}_{\mathrm{a}}=36 \times 10^{-3}$
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