When 9.45g of ClCH₂COOH is added to 500ml. of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH₂ COOH is x * 10^-3. The value of x is__________ (Rounded off to the nearest integer)

$\left[k_{f\left(H_{2}O\right)}=1.86Kkgmo{l}^{-1}\right]$         

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

  $\Delta T_f=i{k}_f.m$           

  $0.5=i*1.86*\frac{9.45}{94.5*500}*1000$         

i = 1.344

Now, using $\alpha =\frac{i-1}{n-1}$  

n for ClCH₂COOH = 2

a =  $\frac{1.344-1}{2-1}=0.344$  

Using

$K_a=\frac{C{\alpha }^2}{1-\alpha }$  

$K_a=\frac{0.2*{\left(0.344\right)}^3}{0.66}=36*10^{-3}$       

So; x = 36