Question

$0.5$ molal aqueous solution of a weak acid (HX) is $20\%$ ionised. If $K_f$ for water is $1.86 \, K \, Kg \, mol^{-1}$, the lowering in freezing point of the solution is

• A. - 1.12 K
• B. 0.56 K
• C. 1.12 K
• D. - 0.56 K

Answer 1

Answer: (C)

$20 \%$ ionisation of weak acid



Total moles at equilibrium $=1-\alpha+\alpha+\alpha=1+\alpha$

$i =\frac{\text { Total moles }}{\text { Initial moles }}=\frac{1+\alpha}{1}=1+0.2=1.2$

We know that, lowcring in freczing point of the solutions $\left(\Delta T _{ b }\right) = i K _{ f } m$

$=1.2 \times 1.86\, K\, kg\, mol ^{-1} \times 0.5$

$=1.116\, K\, \approx 1.12\, K$