When 9.45g of is added to 500 mL of water, its freezing point drops by $0.5°\mathrm{C}$. 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).
$\left[{\mathrm{K}}^{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)=1.86 {\mathrm{Kkgmol}}^{-1}\right]$

${\mathrm{ClCH}}_2\mathrm{COOH}\rightleftharpoons {\mathrm{ClCH}}_2{\mathrm{COO}}^{-}+{\mathrm{H}}^{+}$

i=1+(2-1) $\alpha$
$\mathrm{i}=1+\alpha$
$\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{ik}}_{\mathrm{f}}\mathrm{m}$
$0.5=(1+\alpha )(1.86)\left(\frac{\left(\frac{9.45}{94.5}\right)}{500\times {10}^{-3}}\right) \left(\because m=\frac{no.of moles of solute}{mass of solvent\left(kg\right)}\right)$
$\frac{5}{3.72}=1+\alpha \Rightarrow \alpha =\frac{1.28}{3.72}$
$\alpha =\frac{32}{93}$
 

${\mathrm{ClCH}}_2\mathrm{COOH}\rightleftharpoons {\mathrm{ClCH}}_2{\mathrm{COO}}^{\ominus }+{\mathrm{H}}^{+}$
$\mathrm{C}-\mathrm{C}\alpha$   

${\mathrm{K}}_{\mathrm{a}}=\frac{(\mathrm{C}\alpha {)}^2}{\mathrm{C}-\mathrm{C}\alpha }=\frac{\mathrm{C}{\alpha }^2}{1-\alpha } \mathrm{C}=\frac{0.1}{500/1000}=0.2$
${\mathrm{K}}_{\mathrm{a}}=\frac{0.2(32/93{)}^2}{(1-32/93)}=\frac{0.2\times (32{)}^2}{93\times 61}$
=0.036
$K_{\mathrm{a}}=36\times {10}^{-3}$.

Hence, the required answer is 36.