The freezing point of the solution containing 0.3gm of acetic acid in 30gm of benzene is lowered by $0.42°\mathrm{C}$, then the van't Hoff factor is (${\mathrm{K}}_{\mathrm{f}}$ for benzene = 5.12 $\mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$)

 

Given:

Van't Hoff factor = i

 ${\mathrm{K}}_{\mathrm{f}}$ for benzene = 5.12 $\mathrm{K} \mathrm{kg} {\mathrm{mol}}^{-1}$

Solute = acetic acid $\left({\mathrm{CH}}_3\mathrm{COOH}\right)$

Molecular mass = $12+3\times 1+12+16+16+1=60$

Let observed mass of solute = M

Given mass = 0.3 gm

Solvent = benzene 

Given mass = 30 gm

${\mathrm{\Delta T}}_{\mathrm{f}}=0.42°\mathrm{C}$

$0.42=5.12\times \frac{{\displaystyle \frac{0.3}{\mathrm{M}}}}{{\displaystyle \frac{30}{1000}}}$

$\Rightarrow \mathrm{M}=121.9$

Van't Hoff factor, $\mathrm{i}=\frac{\mathrm{Observed} \mathrm{mass}}{\mathrm{Molecular} \mathrm{mass}}=\frac{60}{121.9}\approx 0.5=\frac{1}{2}$

The van't Hoff factor is $\frac{1}{2}$.