$\begin{array}{1 1}0.945\\0.645\\0.675\\0.975\end{array} $

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Given :

$w=0.2g,W=20g,\Delta T=0.45$

$\Delta T =\large\frac{1000\times K_f\times w}{m\times W}$

$0.45=\large\frac{1000\times 5.12\times 0.2}{20\times m}$

$\therefore m$(observed)=113.78

Now for $2CH_3COOH \leftrightharpoons (CH_3COOH)_2$

$\qquad\quad 1\qquad\qquad\qquad\qquad 0$

$\qquad\quad 1-\alpha\qquad\qquad\qquad \large\frac{\alpha}{2}$

Where $\alpha$ is the degree of association

$\therefore \large\frac{m_{normal}}{m_{observed}}=$$1-\alpha+\large\frac{\alpha}{2}$

$\large\frac{60}{113.78}$$=1-\alpha+\large\frac{\alpha}{2}$

$\Rightarrow \alpha=0.945$

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