Phenol is dimerised as follows
At $t=0$
$$
2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH} \rightleftharpoons\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\right)_2
$$
At equilibrium
$$
(1-\alpha) \quad \alpha / 2
$$
Total moles after dimerisation $=1-\alpha+\alpha / 2=1-\alpha / 2$
$$
\begin{aligned}
\text { Van't Hoff factor (i) } & =\frac{\text { Normal molecular wt. of phenol }}{\text { Abnormal molecular wt. of phenol }}=\frac{94}{m_{a b}} \\
m_{a b} & =\frac{1000 \times K_f \times w}{\Delta T_f \times W}
\end{aligned}
$$
Given that
$$
\begin{aligned}
& K_f=14 \\
& w=72.5 \mathrm{gm} \\
& W=1000 \mathrm{gm} \\
& \therefore \quad m_{a b}=\frac{1000 \times 14 \times 72.5}{7 \times 1000}=145 \\
&
\end{aligned}
$$
$$
\therefore \quad i=\frac{94}{145}
$$

$$
\text { (i) }=\frac{\text { No. of particles after dimerisation }}{\text { No. of particles before dimerisation }}=\frac{1-\alpha / 2}{1}
$$
From eq. (i) and (ii)
$$
\begin{aligned}
1-\alpha / 2 & =\frac{94}{145}=0.65 \\
\frac{\alpha}{2} & =1-0.65=0.35
\end{aligned}
$$
Hence, $35 \%$ phenol is present in dimeric form.