Question

The degree of ionisation of $HF$ in $0.100\, M$ aqueous solution is (freezing point of the solution $=-0.197^{\circ} C$ and $K_{f}$ for water $=186^{\circ}C)$

• A. 0.06
• B. 0.12
• C. 0.03
• D. 0.09

Answer 1

Answer: (A)

0.06

Answer 2

$\Delta T_{f}=i k_{1} m$
$0.197=i \times 1.86 \times 0.1$
$i=\frac{0.197}{1.86 \times 0.1}=1.059 \approx 1.06$
In case of dissociation, $\alpha=\frac{8-1}{n-1}$
$=\frac{1.06-1}{2-1}=0.06=6 \%$