Question

The vapour pressures of two liquids A and B are 80 mmHg and 60 mm Hg respectively at 25°C. What is the vapour pressure of the solution obtained by mixing 3 moles of A and 2 moles of B?

Answer 1

`P_T = (P_2^0 - P_1^0) x_2 + P_1^0`

1 is B and 2 is A.

`P_1^0 = 60  "mm Hg", P_2^0 = 80  "mm Hg"`

`x_2 = 3/(3 + 2) = 0.6`

P_T = (80 - 60) 0.6 + 60

= 20 × 0.6 + 60 = 72 mm Hg