Question

In 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, the freezing point of the solution will be nearest to :

• A. $-{0.481}^{0}C$
• B. $-{0.360}^{0}C$
• C. $-{0.260}^{0}C$
• D. ${0.35}^{0}C$

Answer 1

The correct option is A $-{0.481}^{0}C$

$\Delta T_f=-i{K}_{f}m$

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

$\alpha =0.3$

$n=2$

$i=1+0.3(2-1)$

$i=1.3$

$\Delta T_f=1.3\times 1.85\times 0.2$

$\Delta T_f=0.481{}^{0}C$

$\therefore T_f^{\ast }=0-0.481=-{0.481}^{o}C$

Hence, the correct option is $\text{A}$