A 0.1 molal aqueous solution of a weak acid is 30 % ionized. If Kf for water is 1.86 °C/m, the freezing point of the solution will be

A 0.1 molal aqueous solution of a weak acid is 30 % ionized. If $K_{f}$ for water is 1.86 °C/m, the freezing point of the solution will be

A
$- {0.180}^{\circ} C$
B
$- {0.54}^{\circ} C$
C
$- {0.36}^{\circ} C$
D
$- {0.24}^{\circ} C$

Solution:

Molality, $m = 0.1$ ; vant Hoff's factor, i for a weak acid $= 1 + α$ where $α =$ degree of dissociation $= 30 %$

$= 30 / 100 = 0.3$

But $i = 1 + α . So, i = 1 + 0 .3 = 1 .3$

Thus, ${ΔT}_{f} = {iK}_{f} \times m = 1 .3 \times 1 .86 \times 0 .1 = 0 {.24}^{\circ} C$

$∴$ Freezing point $= 0 - {0.24}^{\circ} C = - {0.24}^{\circ} C$

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