Question

$KBr$ is $80\%$ dissociated in aqueous solution of $0.5m$ concentration (given, $K_f$ for water $=1.86KKgmo{l}^{-1})$. The solution freezes at

• A. 271.326 K
• B. 272 K
• C. 270.5 K
• D. 268.5 K

Answer 1

Answer: (A)

$\Delta T_f=i\times k_f\times m$

Here; degree of dissociation $=80\%$

$\because i=1-\alpha +n\alpha \rightleftharpoons K^{+}+B{r}^{-}$

$\therefore i=\frac{1-0.8+2\times 0.8}{1}=1.8$

$\therefore \Delta T_f=1.8\times 1.86\times 0.5$

$\Rightarrow \Delta T_f=1.674k$

$\Delta T_f=T_f^0-T_f\Rightarrow 1.674=273-T_f$

$T_f=\mathrm{273.1.674}\Rightarrow T_f=271.326K$