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

Options

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.

Answer

A, B

Explanation

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`