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. -0.480$^\circ$C
• B. -0.360$^\circ$C
• C. -0.260$^\circ$C
• D. + 0.480$^\circ$C

Answer 1

Answer: (A)

$HX \rightleftharpoons H ^{+}+ X ^{-}$
$i=\frac{1-0.3+0.3+0.3}{1}=1.3$
$\Delta T_f=i K_f m$
$=1.3 \times 1.85 \times 0.2=0.48$
$\therefore T_f=(0-0.48)^{\circ} C =0-48^{\circ} C$.