$Initial\; moles\quad \quad 1 \quad \quad \quad \quad 0$

$at\; equilibrium \quad 1-\alpha\quad \quad \alpha / 2$

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 $=94.5 \%$