Given :
$w=0.2$g
$W=20g$
$\Delta T=0.45$
Applying,$\Delta T=\large\frac{1000\times K_f\times w}{m\times W}$
Or,$0.45=\large\frac{1000\times 5.12\times 0.2}{20\times m}$
$\therefore m$(observed)=113.78(acetic acid)
$2CH_3COOH\qquad\leftrightharpoons\qquad (CH_3COOH)_2$
Before dissociation $1\qquad 0$
After dissociation $1-\alpha\qquad \large\frac{\alpha}{2}$
$\alpha$-Degree of dissociation
Molecular weight of acetic acid =60
$i=\large\frac{\text{normal molecular mass}}{\text{observed molecular mass}}$
$\therefore \large\frac{m_{normal}}{m_{observed}}=$$1-\alpha +\large\frac{\alpha}{2}$
$\large\frac{60}{113.78}=$$1-\alpha+\large\frac{\alpha}{2}$
$\therefore \alpha=0.945$