Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App