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Two liquids A and B form ideal solutions...

Two liquids A and B form ideal solutions. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm 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 Hg. The vapour pressures of A and B in their pure states are respectively

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Let the vapour pressure of pure `A be = P_(A)^@`, and the vapour pressure of pure `B be = P_(B)^@`.
Total vapour pressure of solution (1 mol A + 3 mol B )
`= chi_(A)P_(A)^@ + chi_(B)P_(B)^@` [`chi_(A)` is mole fraction of A and `chi_(B)` is mole fraction of B]
`500 = 1/4p_(A)^@ + 500 = 3/4p_(B)^@`
or `2000 =p_(A)^@ + 3p_(B)^@`
Total vapour pressure of solution (1 mol A + 4 mol B)
`=1/5 p_(A)^@ +4/5 p_(B)^@`
`510 = 1/5 p_(A)^@ + 4/5 p_(B)^@`
`2250=p_(A)^@ + 4p_(B)^@ `
Solving Eqs. (i) and (ii), we get
`p_(B)^@ =550` mm Hg = Vapour pressure of pure B
`p_(A)^@ =350` mm Hg = Vapour pressure of pure A
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