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.

Answer

`P_(A)^(0)=400mm` of `Hg, P_(B)^(0)=600mm` of `Hg`

Explanation

In first case, `X_(A)=(1)/(4),X_(B)=(2)/(4) & P_(T)=550`
According to Raoult's law,
`X_(A)P_(A)^(0)+ X_(B)P_(B)^(0)=P_(T)`
so `(1)/(4)P_(A)^(0)A+(3)/(4)P_(B)^(0)=550 "or" P_(A)^(0)+3P_(B)^(0)=2200` ....(i)
In second case
`X_(A)^(l)=1//5,X_(B)^(l)=4//5 & P_(T)=560`
`therefore (1)/(5)P_(A)^(0)+(4)/(5)P_(B)^(0)=560 "or" P_(A)^(0)+4P_(B)^(0)=2800` ....(ii)
Subtracting (i) from (ii), we get
`P_(A)^(0)=400 mm P_(B)^(0)=600 mm` of `Hg`.