Question

A mixture of two volatile liquids a and b for 1 and 3 moles respectively has a vapour pressure of 300 mm at 27 degree celsius if one mole of a further added to the solution the vapour pressure become 290 mm at 27 degree celsius the vapour pressure of pure a is

Answer 1

Given:

1. Moles of A = 1

2. Moles of B = 3

3. Vapor pressure = 300 mm

4. New Vapor pressure upon adding one mole of A = 290 mm

To find:

Vapor pressure of pure A

Solution:

• We know that, P = Xa x Pa + Xb x Pb

                            300 = 1 / (1+3) Pa + 3 / (1+3) Pb

                            300 = 0.25Pa + 0.75Pb                            eq(1)

• Now, when 1 mole of A is added, then,

       290 = 2 / (2+3) Pa + 3 / (2+3) Pb

       290 = 0.4Pa + 0.6Pb                                                     eq(2)

• On solving eq(1) and eq(2), we get,

        Pa = 250mm

• Thus, the vapor pressure of pure A is 250 mm.