Question

A solution contains $3.22g$ of $HCl{O}_2$ in $47.0g$ of water. The freezing point of the solution is $271.10K$.
Calculate the fraction of $HCl{O}_2$ that undergoes dissociation to $H^{+}$ and $Cl{O}_2^{-}$.
Given: $K_f\left(water\right)=1.86Kkgmo{l}^{-1}$.

• A. $\alpha =0.102$
• B. $\alpha =0.204$
• C. $\alpha =0.30$
• D. $\alpha =0.60$

Answer 1

The correct option is A $\alpha =0.102$

The molar mass of chlorous acid $HCl{O}_2$ is $68.5$ g/mol.

$3.22$ g chlorous acid $=\frac{3.22}{68.5}=0.047$ moles. The mass of water is $47.0$ g or $0.047$ kg (as $1$ kg $=1000$ g).

The molality of the solution is the ratio of the number of moles of chlorous acid to the mass of water (in kg) $m=\frac{0.047}{0.047}=1$

The depression in the freezing point of the solution $\Delta T_f=273.15-271.10={2.05}^{o}C$

The vant Hoff factor $i=\frac{\Delta T_f}{K_{f}m}$

$i=\frac{2.05}{1.86\times 1}=1.102$

The fraction of $HCl{O}_2$ that undergoes dissociation $\alpha =\frac{i-1}{n^{\prime }-1}$

$\alpha =\frac{1.102-1}{2-1}=0.102$