Determine the vapour pressure when 2.20 mol of A and 4.30 mol of B are mixed. What is the composition of the vapour phase?

Question

A mixture of liquids A and B forms an ideal solution. The vapour pressure of pure

liquids A and B are 8000 Pa and 3600 Pa respectively at 350K.

i) Determine the vapour pressure when 2.20 mol of A and 4.30 mol of B are

mixed. What is the composition of the vapour phase?

Answer 1

First, we'll assume both A and B are volatile liquids, and then we'll apply Dalton's law of partial pressure and Raoult's law.

We will want to find the mole fraction for each liquid.

Total moles = 2.20 mol + 4.30 mols = 6.50 mols

Mol fraction A (X_A) = 2.20 / 6.50 = 0.338

Mol fraction B (X_B) = 4.30 / 6.50 = 0.662

Next, we use the mol fractions to determine the vapor pressure of each in the vapor phase:

P_A = 0.338 x 8000 Pa = 2704 Pa

P_B = 0.662 x 3600 Pa = 2383 Pa

Vapor pressure of mixture = 2704 Pa + 2383 Pa = 5087 Pa

Composition of the vapor phase is determined from the vapor pressure of each:

X_A in vapor phase = 2704 Pa / 5087 Pa = 0.53

X_B in vapor phase = 2383 Pa / 5087 Pa = 0.47

So, the composition of the vapor phase is 53% A and 47% B