Question

A $0.1$ molal aqueous solution of a week acid is $30$% ionised. If $K_f$ for water is ${1.86}^{\circ }C/m$, the freezing point of the solution will be:

• A. $-{0.18}^{\circ }C$
• B. $-{0.54}^{\circ }C$
• C. $-{0.36}^{\circ }C$
• D. $-{0.24}^{\circ }C$

Answer 1

The correct option is D $-{0.24}^{\circ }C$

Freezing point depression $\left(△T_f\right)=i{K}_{f}m$

$\underset{\underset{1-0.3}{1-\alpha }}{HA\to }\underset{\underset{0.3}{\alpha }}{H^{+}}+\underset{\underset{0.3}{\alpha }}{A^{-}}$

$i=1-0.3+0.3+0.3$

$i=1.3$

$\therefore △T_f=1.3\times 1.86\times 0.1={0.2418}^{\circ }C$

$T_f=0-{0.2418}^{\circ }C$

$=-{0.2418}^{\circ }C$

Hence, option D is correct.