Question

Benzene and toluene form nearly ideal solution .At 313K the vapour pressure of pure benzene is 150mm Hg and of pure toluene is 50 mm Hg .calculate vapour pressure of mixture of these two containing their equal masses at 313 K .

Answer 1

In solution we have equal masses of benzene and toluene are present

Let mass of benzene be wb

and mass of toulene be wt

But the mass of Benzene = mass of Toulene = wb =wt

Number of moles of benzene=wb/78 (molar massof benzene is 78)

Number of moles of toluene=wt/92 (molar massof toulene is 92) = wb/92 ( since wb =wt)

Hence,

Mole fraction of benzene ,Xb=Number of moles of benzene /Total number of moles

=wb/78/(wb/78+wb/92)

=wb/78/(wb/78+wb/92)

= (wb/78 )/ (78wb+92wb /78x 92)

since its division of fractions we multiply by the reciprocal of the denominator fraction
= (wb/78) x ( 78 x 92 /170 wb)

= 0.54

Hence mole fraction of benzene =Xb= 0.54

Mole fraction of toulene = Xt =1-0.54=0.46

Hence

If P is total pressure of this mixture

Then

P = Xb *Pb + Xt *Pt = 0.54 x150 + 0.46 x 50 = 104mmHg