In a 0.2 molal aqueous solution of a weak acid HX, the degree of dissociation is 0.3. Taking Kf for water as 1.85 K kg mol−1, the freezing point of the solution will be nearest to:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −0.480 °C
Depression of freezing point is a colligative property, so the dissociation of HX is counted through the van’t Hoff factor i:
\( \displaystyle \Delta T_f=i\,K_f\,m \)
Dissociation: \( \ce{HX} \rightleftharpoons \ce{H^+ + X^-} \); for each mole of HX taken, (1 − α) mol stays as HX and α mol each of \( \ce{H^+} \) and \( \ce{X^-} \) form.
Total particles: (1 − α) + α + α = 1 + α, so i = 1 + 0.3 = 1.3.
\( \displaystyle \Delta T_f=1.3\times1.85\,\mathrm{K\,kg\,mol^{-1}}\times0.2\,\mathrm{mol\,kg^{-1}} \)
\( \displaystyle \Delta T_f=0.481\,\mathrm{K} \)
Water freezes at 0 °C, and a change of 0.481 K is a change of 0.481 °C, so the solution freezes at 0 − 0.481 = −0.481 °C ≈ −0.480 °C.