When 12.2g of benzoic acid is dissolved in 100g of water, the freezing point of solution was found to be -0.93°C $\left[K_f(H_{2}O=1.86Kkgmo{l}^{-1}\right].$  The number (n) of benzoic acid molecules associated (assuming 100% association) is…………,

0.5 % KCl solution has molality (m) = $\frac{0.5\times 1000}{74.5\times 99.5}$  

$KC{l}_{\left(aq\right)}?K_{\left(ag\right)}^{+}+C{l}_{\left(aq\right)}^{-}$  

1 - a        a          a

And I =  $\left(1-\cancel{\alpha }+\cancel{\alpha }+\alpha \right)=1+\alpha$  

$i=\frac{\Delta Tf}{kf\times m}=\frac{0.24\times 74.5\times 99.5}{1.8\times 0.5\times 1000}=1+\alpha$  

1.976 = 1 + a

$\therefore \alpha =0.976$  

% = 97.6%

the nearest 98.

$\Delta T_f=0.5°C$

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 

  $\Delta T_b=i{K}_{b}m$ ${}_{}{}_{}$

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

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%