Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If ${K_f}$ of water is ${1.86^ \circ }$ c/m, the freezing point of solution will be:
A. -${0.24^ \circ }$ C
B. -${0.18^ \circ }$ C
C. -${0.54^ \circ }$ C
D. -${0.36^ \circ }$ C

Answer 1

Hint:
${K_f}$ (molal freezing point depression constant) of water is 1.86 °C/m when moles of solute per kilogram of water are unity.

Complete Step By Step Solution:
Let us first look at the relationship between the depression in the freezing point and the molality of the solution :
$\Delta {T_f}$ = $i$ ${K_f}$ $m$
Let $\alpha $ be the degree of dissociation of the weak acid.
The dissociation equilibrium of the weak acid is given by
$HA \to {H^ + } + {A^ - }$
$1-\alpha \; \;\; \alpha \; \;\;\;\;\;\;\; \alpha $

And it is given that $\alpha $=0.3
Therefore it becomes
$HA \to {H^ + } + {A^ - }$
$1-\alpha \;\;\; \alpha \;\;\;\; \;\;\;\; \alpha $
1-0.3 $\;$ 0.3 $\;\;\;\;\;\;$ 0.3

The Van’t Hoff factor $i$ is the total no. of ions or molecules given by the dissociation of 1 molecule of solute.
$i$ = 1-0.3 + 0.3 + 0.3
$i$=1.3

Substituting the values in the expression
$\Delta {T_f}$= 1.3$ \times $ 1.86$ \times $0.1
$\Delta {T_f}$=0.2418

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

${T_f}$=${0.2418^ \circ }$C
Hence the freezing point of the solution will be -${0.24^ \circ }$ C

Hence the correct option is 1. -${0.24^ \circ }$ C

Note:
${K_f}$is a constant or a number which is always the same. The value of ${K_f}$depends only on the type of solvent and not on the type of solute added.