Question

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
(a) – 0.360ºC (b) – 0.260ºC
(c) + 0.481ºC (d) – 0.481ºC

Answer 1

answer : option (d) - 0.481°C

molality of given aqueous solution, m =0.2 molal

degree of ionisation, α = 0.3

Kf for water = 1.85

dissociation reaction will be .....

HX ⇒ H^+ + X-

so, i = 1- α + α + α= 1 + α

⇒i = 1 + 0.3 = 1.3

now using formula, ∆T_f = i × m × K_f

= 1.3 ×0.2 × 1.85

= 0.26 × 1.85

= 0.481

so, ∆T_f = +0.481°C = 0° - T_f

⇒T_f = -0.481 °C

hence, option (d) is correct choice.

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