Question

In a $0.2$ molal aqueous solution of a weak acid, $HX$. the degree of ionization is $0.3$. If $K_f$ for water is $1.85$, find the freezing point of the solution.

• 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 $-{0.481}^{o}C$

We know,

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

$\alpha =0.3,n=2$ (since two ions are created, i.e., $H^{+}and{X}^{-}$)

Putting these values in the equation,

$i=1+0\cdot 3(2-1)$

$i=1\cdot 3$

Again,

$\Delta T_f=K_f\times m\times i$

$K_f=1.85,m=0.2$

$\Delta T_f=1\cdot 85\times 2\times 1\cdot 3$

So,

$T_f=0-0.481=-0\cdot {481}^{o}C$

Option D is correct.