Question

$KBr$ is $80 \%$ dissociated in aqueous solution of $0.5 \,m$ concentration (given, $K_{f}$ for water $\left.=1.86 \,K \,Kg\, mol ^{-1}\right)$. 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 $=80 \%$

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

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

$\therefore \Delta T _{ f }=1.8 \times 1.86 \times 0.5$

$\Rightarrow \Delta T _{ f }=1.674 k$

$\Delta T _{ f }= T _{ f }^{0}- T _{ f } \Rightarrow 1.674=273- T _{ f }$

$T _{ f }=273.1 .674 \Rightarrow T _{ f }=271.326 K$