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At 300 K, the vapour pressure of an idea...

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.

A

`300 Hg`

B

`600 Hg`

C

`400 Hg`

D

`500 Hg`

Text Solution

Verified by Experts

The correct Answer is:
3
First situtation
`chi_(A)=n_(A)/n_("total")=1/(1+3)=1/4`
`chi_(B)=n_(B)/n_("total")=3/(1+3)=3/4`
Applying `1^(st)` Raoult's law, we have
`P_("total")=P_(A)^(0) chi_(A)+P_(B)^(0)chi_(B)`
`550 mmHg=1/4 P_(A)^(0)+3/4P_(B)^(0) …(1)`
Second situation
When one more of `B` is added to the solution:
`chi_(A)=1/(1+4)=1/5`
`chi_(B)=4/(1+4)=4/5`
Applying 1st Raoult's law we get
`P_("total")=P_(A)^(0) chi_(A)+P_(B)^(0)chi_(B)`
`(550+10) mmHg=1/5 P_(A)^(0)+4/5 P_(B)^(0) ...(2)`
Solving Eqs (1) and (2) simultaneously, we get `P_(A)^(0)=400 mmHg`
`P_(B)^(0)=600 mmHg`
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