Question

60
53. Benzene (C H, 78 g/mol) and toluene (C,Hg,
92 g/mol) form an ideal solution. At 60°C the
vapour pressure of pure benzene and pure
toluene are 0.507 atm and 0.184, respectively.
The mole fraction of benzene in a solution of
these two chemicals that has a vapour pressure
of 0.350 atm at 60°C
(1) 0.514 (2) 0.690 (3) 0.486 (4) 0.190​

Answer 1

Let vapour pressure of pure benzene be Pᵃ = 0.507 and that of pure toluene be Pᵇ = 0.184

Vapour Pressure = P = 0.350

According to Raoult's law,

$P = P^a \chi^a + P^b\chi^b$

We know that,

$\chi^a + \chi^b = 1\\\\\chi^b = 1 - \chi^a$

Substituting this in the equation

$P = P^a \chi^a + P^b(1-\chi^a)$

0.350 = 0.507 (Xᵃ) + 0.184 (1 - Xᵃ)

0.166 = 0.323(Xᵃ)

Xᵃ = 0.5139

   ≈ 0.514