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

Mole fraction of A `=(1)/(1+3)=(1)/(4)`
Mole fraction of B `=(3)/(1+3)=(3)/(4)`
Applying Eqn. 3 we get,
`550=(1)/(4)xxp_(A)^(0)+(3)/(4)p_(B)^(0)`…………..(1)
When one more mole of B was added to the solution,
Mole fraction of A `=(1)/(1+4)=(1)/(5)`
Mole fraction of B `=(4)/(1+4)=(4)/(5)`
Again applying Eqn. (3) , we get
`550+10=(1)/(5)p_(A)^(0)+(4)/(5)p_(B)^(0)` ................(2)
From (1) and (2), we get
`p_(A)^(0)=440mm`
`p_(B)^(0)=600mm`