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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 dimerises in benzene.$K_f$ for benzene =$5.12Kmol^{-1}Kg$

$\begin{array}{1 1}0.945\\0.845\\0.645\\0.545\end{array} $

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Given :
$\Delta T=0.45$
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}$
$\therefore \alpha=0.945$
answered Aug 6, 2014 by sreemathi.v

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