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1 mol benzene (P^(@)("benzene")=42 mm) ...

1 mol benzene `(P^(@)_("benzene")=42 mm)` and 2 mol toluence `(P^(@)_("toluene")=36 mm)` will have

A

Total vapour pressure of `38 mm`.

B

Mole fraction of vapour of benzene above liquid mixture is `7//19`.

C

Positive devaition from Raoult's law.

D

Negative devaition from Raoult's law.

Text Solution

Verified by Experts

The correct Answer is:
A, B
Benzene and toluene `rArr` Ideal solutions.
`P_("Total")= P_(A)^(@)chi_(A) + P_(B)^(@)chi_(B) = 42 xx 1/3 + 36 xx 2/3=38 mm Hg`
`chi^(V)_("Benzene")=P_("Benzene")/P_("Total") = (42 xx 1/3)/38=7/19`
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