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If Vapour pressure of pure liquids 'A' ...

If Vapour pressure of pure liquids 'A' & 'B' are 300 and 800 torr respectively at `25^@C`. When these two liquids are mixed at this temperature to from a solution in which mole percentage of 'B' is 92, then the total vapour pressure is observed to be `0.95` atm. Which of the following is true for this solution.

A

`DeltaV_("mix") gt 0`

B

`DeltaH_("mix") lt 0`

C

`DeltaV_("mix")=0`

D

`DeltaH_("mix")=0`

Text Solution

Verified by Experts

The correct Answer is:
B

According to Raoult's law
`P_(T)=(0.08xx300+0.92xx800)` torr
`=(24+736)` torr
= 760 torr =1 atm
`P_(exp)=0.95 "atm" lt 1 "atm"`
Hence the solution shows `-ve` deviation
so `DeltaH_("mix")lt 0 ,and DeltaV_("mix") lt 0`
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