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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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The freezing point of 0.02mole fraction solution of acetic acid in benzene is 277.4K.Assuming molality equal to molarity,calculate $K_c$ for the equilibrium reaction $2CH_3COOH \;\;\leftrightharpoons\;\;(CH_3COOH)_2$.

$\begin{array}{1 1}2.787M^{-1}\\3.787M^{-1}\\4.787M^{-1}\\5.787M^{-1}\end{array} $

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Given :
$T_f$(benzene)=278.4K
$K_f$(benzene)=$5.0kg mol^{-1}$
From the expression ,$-\Delta T_f=K_fm$
We get
$m=\large\frac{\Delta T_f}{K_f}=\frac{(278.4-277.4)}{5.0}$
$\Rightarrow 0.2mol kg^{-1}$
(ie) $C=0.2mol dm^{-3}$
From the given mole fraction ,we can calculate the initial concentration of acetic acid.We have
$x_2=\large\frac{n_2}{n_1+n_2}$$\simeq \large\frac{n_2}{n_1}=\frac{n_2}{(w_1/M_1)}$
$\Rightarrow \large\frac{n_2}{w_1}$$\times M_1$
$\Rightarrow \large\frac{w_2}{M_2\times w_1}$$\times M_2=m_0M_1$
$m_0=\large\frac{x_2}{M_1}=\frac{0.02}{78\times 10^{-3}}$
$\Rightarrow 0.257mol kg^{-1}$
(ie) $C_0=0.257mol dm^{-3}$
Now for the equilibrium reaction ,
$2CH_3COOH \;\;\leftrightharpoons\;\;(CH_3COOH)_2$.
$C_0-2x\qquad\qquad\qquad x$
We get,$C=C_0-2x+x=C_0-x$
$0.2M=0.257M-x$
Or,$x=0.057M$
Thus $K_c=\large\frac{[(CH_3COOH)_2]}{[CH_3COOH]}$
$\Rightarrow \large\frac{x}{(C_0-2x)^2}$
$\Rightarrow \large\frac{0.057}{(0.257-2\times 0.057)^2}$
$\Rightarrow 2.787 M^{-1}$
answered Aug 7, 2014 by sreemathi.v
 

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