Two liquids A and B form ideal solution. At `300 K`, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is `550 mm` of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by `10 mm` of Hg. Determine the vapour pressure of a and B in their pure states.

Explanation

Let the vapour pressure of pure A be `= p_(A)^(0)` and the vapour pressure of pure B be `= p_(B)^(0)`
Total vapour pressure of solution (1 mole A + 3 mole B)
`= x_(A).p_(A)^(0)+x_(B).p_(B)^(0)[x_(A)" si mole fraction of A and "x_(B)"is mole fraction of B"]`
`550=(1)/(4)p_(A)^(-)+(3)/(4)p_(B)^(0)`
or`2200=p_(A)^(0)+3p_(B)^(0)`....(i)
Total vapour pressure of solution (1 mole A + 4 mole B)
`= (1)/(5)p_(A)^(0)+(4)/(5)p_(B)^(0)`
`560=(1)/(5)p_(A)^(0)+(4)/(5)p_(B)^(0)`
or `2800=p_(A)^(0)+4p_(B)^(0)`....(ii)
Solving eqs. (i) and (ii),
`p_(B)^(0)=600` mm of Hg = vapour pressure of pure B
`p_(A)^(0)=400` mm Hg = vapour pressure of pure A.