A 0.5 percent solution of potassium choride was found to freeze at -0.24°C. The percentage dissociation of potassium chloride is__________. (Nearest integer)

(Molal depression constant for water is 1.80 K kg mol^-1 and molar mass of KCl is 74.6 g mol^-1)

$\Delta T_f=0.5°C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

$\Delta T_f=i{k}_f.m$           

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

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

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

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

Kindly consider the following Image 

$\Delta T_f=K_f\times m\times i$

0.93 = 1.86 × 1 × i

i = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ $i=1+\left(\frac{1}{n}-1\right)$ $\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$\therefore n=2$

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

$m=\frac{w\times 1000}{M\times W_{solvent}}$

For acetone solution,

$0.17=1\times 1.7\times \frac{1.22\times 1000}{M\times 100}$

For Benzene solution,

$2Acid\to {\left(Acid\right)}_2\therefore i=1/2$

$\Delta T_b=i\times K_b\times m$

$=\frac{1}{2}\times 2.6\times \frac{1.22\times 1000}{122\times 100}°C$

= $0.13°C=13\times 10^{-2}°C$ ${}^{}$

$\therefore x\times 10^{-2}=13\times 10^{-2}$

$\therefore x=13$

$t_{1/2}=200days=\frac{0.693}{k}$

$t=\frac{2.303}{k}lo{g}_{10}\frac{N_0}{N}$

83 =  $\frac{2.303}{0.693}\times 200log\frac{N_0}{N}$

$83=\frac{200}{0.3010}log\frac{N_0}{N}$

0.125 = $log\frac{N_0}{N}$

$\frac{N_0}{N}=antilog\left(0.125\right)$

= 1.333

Activating remaining = $\frac{N}{N_0}\times 100$

$=\frac{\frac{N_0}{1.333}}{N_0}\times 100$

= 75%