Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. 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. $-{0.24}^{\circ }C$
• D. $-{0.36}^{\circ }C$

Answer 1

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

The relationship between the depression in the freezing point and the molality of the solution is as given below.
$\Delta T_f=i{K}_{f}m$

Let α be the degree of dissociation of the weak acid.
The dissociation equilibrium of the weak acid is as represented below,

The Van't Hoff factor 'i' is the total number of ions/molecules given by the dissociation of 1 molecule of solute.

$i=1-0.3+0.3+0.3i=1.3$

$\Delta T_f=i\times K_f\times m=1.3\times 1.86\times 0.1=0.2418$

$T_f=0-0.2418=-{0.2418}^{o}C$

Hence, the freezing point of the solution will be $-{0.24}^{o}C$