A mixture of two volatile liquids a and b for 1 and 3 moles respective

1.	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» <P>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.

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»

<P>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:

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

300 = 0.25Pa + 0.75Pb eq(1)

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

290 = 0.4Pa + 0.6Pb eq(2)

Pa = 250mm