\(75.2 \mathrm{~g}\) of \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}\) (phenol) is dissolved in a solvent of \(K_f=14\). If the depression in freezing point is \(7 \mathrm{~K}\), then find the \(\%\) of phenol that dimerises.

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 \(=1-\alpha+\alpha / 2=1-\alpha / 2\)
\(\text {Van't Hoff factor (i)} =\) \(\frac{\text {Normal molecular wt. of phenol}}{\text {Abnormal molecular wt. of phenol}}\)\(=\frac{94}{m_{a b}} \)
\(m_{a b} =\frac{1000 \times K_f \times w}{\Delta T_f \times W}\)
Given that
\(K_f=14 \)
\(w=72.5 \mathrm{gm} \)
\(W=1000 \mathrm{gm} \)
\(\therefore m_{a b}=\frac{1000 \times 14 \times 72.5}{7 \times 1000}=145\)
\(\therefore \quad i=\frac{94}{145}\)
\(\text {(i)}=\frac{\text {No. of particles after dimerisation}}{\text {No. of particles before dimerisation}}\)\(=\frac{1-\alpha / 2}{1}\)
From eq. (i) and (ii)
\(1-\alpha / 2 =\frac{94}{145}=0.65 \)
\(\frac{\alpha}{2} =1-0.65=0.35\)
Hence, \(35 \%\) phenol is present in dimeric form.