In a 0.2 molal aqueous solution of a weak acid HX

In a 0.2 molal aqueous solution of a weak acid HX, the degree of ionization is 0.3. Taking Kf for water as 1.85, the freezing point of the solution will be nearest to

  1. – 0.480 °C
  2. – 0.360 °C
  3. – 0.260 °C
  4. + 0.480 °C

Answer

HX ↔ H+ + X–

1        0       0

1–0.3  0.3 0.3

Total number of moles after dissociation = 1 – 0.3 + 0.3 + 0.3 = 1.3

(Kf (observed))/(Kf (experimental)) = (no. of moles after dissociation)/(no. of moles before dissociation)

 Or

 Kf (observed)/1.85 = 1.3/1

 Kf (observed) = 1.85 × 1.3 = 2.405

ΔTf = Kf × molality = 2.405 × 0.2 = 0.4810

Freezing point of solution = 0 – 0.481 = – 0.481°C

The correct option is A.