Question

For a 0.2 molal aqueous solution of weak acid hx (the degree of dissociation = 0.3) the freezing point (in ºc) is (given kf=1.85kf=1.85ºc molal−1molal−1)

Answer 1

Explanation:

.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