Question

$75.2 \,g$ of $C _{6} H _{5} OH$ (phenol) is dissolved in a solvent of $K_{f}=14$. If the depression in freezing point is $7\, K$ then find the percentage of phenol that dimerises.

Answer 1

The dimerization reaction of phenol is
$2 C _{6} H _{5} OH \rightleftharpoons\left( C _{6} H _{5} OH \right)_{2}$

Here, $C$ is the concentration of phenol:
$C =\frac{75.2}{ Mol . w t}=\frac{75.2}{94}=0.8$
and $\alpha$ is the degree of dissociation.
Then $i = C - Ca + Ca / 2=\frac{C(2-\alpha)}{2}$
Freezing point depression is given by
$\Delta T_{f}=K_{f} \times i $
$7=14 \times 0.8 \times\left(2-\frac{\alpha}{2}\right)$
$\alpha=0.75=75 \%$