d
\( \quad \quad \quad \quad 2 \mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{2} \)

\(Initial\; moles\quad \quad 1 \quad \quad \quad \quad 0\)

\(at\; equilibrium \quad 1-\alpha\quad \quad \alpha / 2\)

Total number of moles at equilibrium \(=1-\alpha+\alpha / 2 \Rightarrow i=1-\alpha / 2(\text { association })\)

\(\Delta T_{\mathrm{r}}=K_{\mathrm{r}} \times m \times i\)

\(0.45=(5.12) \times \frac{(0.2 / 60)}{(20 / 1000)} \times\left(1-\frac{\alpha}{2}\right)\)

\(\left(1-\frac{\alpha}{2}\right)=0.527 \Rightarrow \alpha=0.945\)

So, the percentage of acetic acid that dimerizes \(=94.5 \%\)