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The freezing point of 0.02 mole fraction solution of acetic acid in benzene is 277.4K.Assuming molality equal to molarity,calculate $K_c$ for the equilibrium reaction $2CH_3COOH\;\;\leftrightharpoons\;\;(CH_3COOH)_2$.Given : $T_f$(benzene) =$278.4$ and $K_f$(benzene)=5.0K kg $mol^{-1}$.

$\begin{array}{1 1}2.794M^{-1}\\3.994M^{-1}\\4.794M^{-1}\\3.794M^{-1}\end{array} $

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From the expression,$-\Delta T_f=K_fm$
We get,
$m=\large\frac{-\Delta T_f}{K_f}$
$\;\;=0.2 mol kg^{-1}$
(ie) $C=0.2mol dm^{-3}$
Molality =molarity
Now $x_2=\large\frac{n_2}{n_1+n_2}$$\approx \large\frac{n_2}{n_1}$
$\Rightarrow \large\frac{n_2}{(w_1/M_1)}$
$\Rightarrow \large\frac{n_2}{w_1}$$(M_1)$
$\Rightarrow m_0M_1$
$m_0=\large\frac{x_2}{M_1}=\frac{0.02}{78\times 10^{-3}}$
$\;\;\;\;\;=0.257molkg ^{-1}$
$C_0=0.257mol dm^{-3}$
Now for the equilibrium constant
$\;\;\;C_0-2x\qquad\qquad x$
We get,$C=C_0-2x$
$0.2M=0.257M-x$ or $ x=0.057M$
Thus $K_c=\large\frac{[(CH_3COOH)_2]}{[CH_3COOH]^2}$
$\Rightarrow \large\frac{x}{(C_0-2x)^2}$
$\Rightarrow \large\frac{0.057}{(0.257-2\times 0.057)^2}$
$\Rightarrow 2.794M^{-1}$
answered Aug 8, 2014 by sreemathi.v

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