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 $1.85$, the freezing point of the solution will be nearest to:

• A. $-0.360$
• B. $-0.260$
• C. $+0.480$
• D. $-0.481$

Answer 1

Answer: (D)

$-0.481$

$\text{Molality,}m=0.2m$

$\alpha =0.3;\alpha =\frac{i-1}{n-1}\lceil HX⟶H^{+}+X^{-};n=2\rceil$

$K_f=1.85Kkg/mol;T_f=?$

$0.3=\frac{i-1}{2-1}\Rightarrow i=1.3$

${\Delta T}_f={iK}_{f}m$

${T_f}^o-T_f=1.3\times 1.85\times 0.2$

$0^{o}C-T_f={0.481}^{o}C$

$T_f=-{0.481}^{o}C$