Question

$0.5$ molal aqueous solution of a weak acid $\left(HX\right)$ is $20\%$ ionized. If $K_f$ for the water is $1.86Kkgmo{l}^{-1}$, the lowering in freezing point of the solution is:

• A. $-0.56K$
• B. $-1.12K$
• C. $0.56K$
• D. $1.12K$

Answer 1

The correct option is C $0.56K$

Freezing point depression, $\Delta T_F$

$\Delta T_F=K_F.b.i$

$K_F$ - cryoscopic constant

$b$ - molality

$i$ - van't hoff factor

given -

$HX\to H^{+}+X^{-}$

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

$\alpha$ - degree of dissociation

$n$ - number of ion

$\alpha =\frac{20}{100}=0.2$

$n=2$

$i=0.2\times 2+(1-0.2)=1.2$

$\Delta T_F=1.86\times 0.5\times 1.2=1.12K$

So, option (d) is the correct answer.