From the expression,$-\Delta T_f=K_fm$
We get,
$m=\large\frac{-\Delta T_f}{K_f}$
$\;\;\;=\large\frac{(278.4-277.4)}{5.0}$
$\;\;=0.2 mol kg^{-1}$
(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
$2CH_3COOH\;\;\leftrightharpoons\;\;(CH_3COOH)_2$
$\;\;\;C_0-2x\qquad\qquad x$
We get,$C=C_0-2x$
$0.2M=0.257M-x$ or $ x=0.057M$
Thus $K_c=\large\frac{[(CH_3COOH)_2]}{[CH_3COOH]^2}$
$\Rightarrow \large\frac{x}{(C_0-2x)^2}$
$\Rightarrow \large\frac{0.057}{(0.257-2\times 0.057)^2}$
$\Rightarrow 2.794M^{-1}$