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The freezing point of a solution containing 0.2g of acetic acid in 20.0g benzene is lowered by $0.45^{\large\circ}$C.Calculate the degree of association of acetic acid in benzene.Assume acetic acid dimerizes in benzene $K_f$ for benzene is $5.12Kmol^{-1}kg$.

$\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}$
$\Rightarrow \alpha=0.945$
answered Aug 7, 2014 by sreemathi.v

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