Two liquids `A` and `B` form an ideal solution. At `300 K`, the vapour pressure of a solution containing `1 mol` of `A` and `3 mol` fo `B` is `550 mm Hg`. At the same temperature, if `1 mol`more of `B` is added to this solution, the vapour pressure of the solution increases by `10 mm 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)` is mole fraction of `A` and `X_(B)` is mole fraction of `B]`
`550 =(1)/(4) p_(A)^(0) +(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)`
`2800 =p_(A)^(0) +p_(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 of `Hg =` vapour pressure of pure `A`