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Total vapour pressure of mixture of 1 mo...

Total vapour pressure of mixture of `1` mole of volatile component `A(P_(A)^(@)=100 mm Hg)` and `3` mole of volatile component `B(P_(B)^(@)=80 mm Hg)` is `90 mm Hg`. Find out nature of solution and nature entropy of solution

A

85 mm Hg

B

86 mm Hg

C

90 mm Hg

D

92 mm Hg

Text Solution

Verified by Experts

The correct Answer is:
B
`P_("total")P_(A)^(@)+P_(B)^(@)X_(B)`
`=100xx(1)/(4)+80xx(3)/(4)=85`
`X_(A)("vapour") (P_(A)^(@)X_(A))/(P_("total"))=(25)/(85)=(5)/(17)`
`X_(B)` (vapourse) `(12)/(17)`
`P_("total" ("dilutions"))=(12)/(17)`
`=100xx(5)/(17)+80xx(12)/(17)`
`=86 mm Hg`
Hence, (B) is the correct answer.
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