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

Options

A. total vapour pressure 38 mm
B. mole fraction of vapours of benzene above liquid mixture is 7/19
C. positive deviation from Raoult's law
D. negative deviation from Raoult's law

Answer

A, B

Explanation

`p=p_(A)^(0) x_(A)+p_(B)^(0)x_(B) A rarr` Benzene, B `rarr` Toluene
`= 42xx(1)/(3)+36xx(2)/(3)`
`=(114)/(3)=38` mm
Mole fraction of benzene in vapour `= (p_("benzne"))/(p_("total"))=(42//3)/(38)`
`=7//19`]