Question

$0.5$ molal aqueous solution of a weak acid (HX) is $20\%$ ionised. If $K_f$ for water is $1.86{\text{K kg mol}}^{-1}$, the lowering in freezing point of the solution is

• A. $1.12\text{K}$
• B. $0.56\text{K}$
• C. $-0.56\text{K}$
• D. $-1.12\text{K}$

Answer 1

The correct option is A $1.12\text{K}$

$\begin{array}{cccc}HX& \rightleftharpoons & H^{+}& X^{-}\\ 1& & 0& 0\\ 1-\alpha & & \alpha & \alpha \end{array}$
$\therefore i=1-\alpha +\alpha +\alpha =1+\alpha =1+0.2=1.2$
So, the lowering of freezing point of the solution,
$\Delta T_f=i\times K_f\times m=1.2\times 1.86\times 0.5=1.116K\approx 1.12K$