At `300 K`, the vapour pressure of an ideal solution containing one mole of `A` and `3` mole of `B` is `550 mm` of `Hg`. At the same temperature, if one mole of `B` is added to this solution, the vapour pressure of solution increases by `10 mm` of `Hg`. Calculate the `V.P.` of `A` and `B` in their pure state.

Explanation

Intially,`P_M=P_A^@X_A+P_B^@X_B`
`550=P_A^@1/(1+3)+P_B^@3/(1+3)` or
`P_A^@+3P_B^@=2200` ..(i)
When, one mole of B is further added to it,
`P_M=P_A^@X_A+P_B^@X_B`
`56=P_A^@1/(1+4)+P_B^@4/(1+4):.`
`P_A^@+4P_B^@=2800` .(ii)
Solving equations (i) and (ii), we get
`P_A^@=440mm`, `P_B^@=6000mm`