Question

The freezing point of aqueous solution of 0.1 m $H{g}_{2}C{I}_2$ will be:
(If $H{g}_{2}C{I}_2$ is 80% ionised in the solution to give $H{g}_2^{2+}$ and $C{I}^{-}$)

• A. $-0.26K_f$
• B. $-2.6K_f$
• C. $-4.2K_f$
• D. $-0.42K_f$

Answer 1

The correct option is A $-0.26K_f$

$H{g}_{2}C{l}_2\rightleftharpoons H{g}_2^{2+}+2C{l}^{-}$

$\alpha =0.8$

No. of ions$=3$.

$i=1+(n-1)\alpha$

$i=1+2\times 0.8=2.6$

$△T_f=i{K}_{f}m$

$m=0.1m$

$△T_f=T_s-T_{sol}$

$T_s=0^{\circ }C$

$\Rightarrow T_{sol}=-i{K}_{f}m$

$T=-0.26K_f$