1.

At 80^(@)C, the vapour p[ressure of pure liquid A is 520 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a mixture solution of A and B boils at 80^(@)C and 1 atomoshere pressure, the amount of A in the mixture is (1 atm = 760 mm of Hg)

Answer»

60 mol precent
52 mol precent
34 mol PRESENT
48 mol precent.

Solution :According to available information,
`P_(A)^(@)=520 MM Hg, P_(B)^(@)=1000MM Hg`
`P_(A)^(@)X_(A)+P_(B)^(@)X_(B)=760 mm Hg`.
`P_(A)^(@)X_(A)+P_(B)^(@)(1-X_(A))760`
`520 X_(A)+1000(1-X_(A))=760`
`520X_(A)+1000-1000X_(A)=760`
`-480X_(A)=-240`
or `X_(A)=240/480=1/2 "or 50 mol precent".


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