Question

A solution of a non-volatile solute in water freezes at −0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and K_f for water is 1.86 degree/molal. Calculate the vapour pressure of this solution at 298 K.

Answer 1

ΔT_f = 0 − (−0.30) = 0.30°C

∵ ΔT_f = K_f × m

∴ `m = (Delta T_f)/K_f`

= `0.30/1.86`

= 0.161

Thus, the given solution contains 0.161 moles of the solute (n) dissolved in 1000 g (W) of the solvent. 

If the solution is treated to be dilute, we have

`(p^circ - p)/p^circ = n/N`

= `(23.51 - p)/23.51 = 0.161/(1000//18)`

`(23.51 - p)/23.51` = 2.9 × 10⁻³

∴ p = 23.51 − (23.51 × 2.9 × 10⁻³)

= 23.44 mm

Hence, the vapour pressure of the given solution at 298 K is 23.44 mm.