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At 310 K, the vapour pressure of an idea...

At `310 K`, the vapour pressure of an ideal solution containing `2 "moles"` of `A` and `3"moles"` 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 increase by `10 mm of Hg`. Calculate the `V.P` of `A` and `B` in their pure state.

Text Solution

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Initially,`P_(M) = P_(A)^(@).X_(A) + P_(B)^(@).X_(B)`
`550 = P_(A)^(@).((2)/(2+3))+P_(B)^(@).((3)/(2+3))`
or `2P_(A)^(@) + 3P_(B)^(@) = 2750` …(1)
When `1` mole of `B` is further added to it,
`P_(M) = P_(A)^(@).X_(A)+P_(B)^(@).X_(B)`
`560 = P_(A)^(@) ((2)/(2+4)) + P_(B)^(@).((4)/(2+4))`
`:. 2P_(A)^(@) + 4P_(B)^(@) = 3360` ...(2)
By eqs. (1) and (2),
`P_(A)^(@) = 460 mm, P_(B)^(@) = 610 mm`
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