1.22 g of an organic acid is separately dissolved in 100 g of benzene (K_b = 2.6 K kg mol^-1) and 100 g of acetone (K_b = 1.7 K kg mol^-1). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17°C. The increase in boiling point of solution in benzene in °C is x * 0^-2. The value of x is ________. (Nearest integer)

[Atomic mass : C = 12.0, H = 1.0, O = 16.0]

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 

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

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.$

$\therefore n=2$

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

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

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%