Question

The freezing point of a solution containing $0.3gms$ of acetic acid in $30gms$ of benzene is lowered by $0.45K$. Calculate the Van't Hoff factor.
(at. wt. of $C=12,H=1,O=16,K_f$ for benzene $=5.12kkgmol{e}^{-1})$

Answer 1

If $i$ is the Van't Hoff factor then depression in freezing point
$△T_f=\frac{i\times 1000\times K_f\times \omega }{W\times M}$
given $△T_f=0.45K$,
$\omega =0.3g$
$W=30g$
$M$ of acetic acid, $C{H}_{3}COOH=60$
$K_f=5.12$
$\therefore i=\frac{△T_f\times W\times M}{1000\times K_f\times \omega }=\frac{0.45\times 30\times 60}{1000\times 5.12\times 0.3}$
$i=0.5273$
Van't Hoff factor is $0.5273$.