Question

An aqueous solution freezes at 272.4k while pure
water freezes at 273k Determine
i) Molality of soln ii) Bp of the soln iii) Lowering in VP.
Given-:
(Kf=1:26, Kb= 0.512, V. P. of pure water = 23.756 mm)
​

Answer 1

Answer:

0.322 m, 373.164 K

Explanation:

Molarity = Δ T/K .. (1)

Δ T = 273 - 272.4 = 0.6 K

K = 1.86 = Freezing constant

Substituting in (1) we get,

Molarity = 0.322 m

Boiling point of water = 100°C = 373 K

ΔTb = K(b) . m = 0.512 x 0.322 --- where K(b) - molar elevation boiling constant

ΔTb = K(b) . m = 0.164 K

Boiling point of solution = 373 K + 0.164 K = 373.164 K

Lowering in VP = ΔP = X(i) x P0