Question

In a $0.2$ molal aqueous solution of a weak acid $HX$, the degree of ionization is $0.3$. Taking $K_f$ for water as ${1.85}^{o}C{m}^{-1}$, the freezing point of the solution will be nearest to:

• A. $-{0.360}^{o}C$
• B. $-{0.260}^{o}C$
• C. $+{0.480}^{o}C$
• D. $-{0.481}^{o}C$

Answer 1

The correct option is D $-{0.481}^{o}C$

Using formula,

$\Delta T_f=-K_f\times molality\times i$

$HX$ will ionize to two units, hence n=2

$i=1+\alpha (n-1)=1+\alpha =1.3$

molality=0.2

Therefore,
$\Delta T_f=1.85\times 1.3\times 0.2={0.481}^{0}C$

$T_s=-{0.481}^{0}C$

Hence, option $D$ is correct.