Question

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

Answer 1

Given: Temperature = 313 K

Vapour pressure of pure benzene, `P_"benzene"^circ​` = 150 mm Hg

Vapour pressure of pure toluene, `P_"toluene"^circ`​ = 50 mm Hg

Equal masses of benzene and toluene are mixed.

Molar mass of benzene = 78 g/mol

Molar mass of toluene = 92 g/mol

Moles of benzene = `78/78` = 1 mol

Moles of toluene = `78/92` = 0.8478 mol

Total moles = 1 + 0.8478 = 1.8478

Mole fraction of benzene `chi_"benzene" = 1/1.8478` = 0.541

Mole fraction of toluene `chi_"toluene" = 0.8478/1.8478` = 0.459

By using Raoult’s law

`P_"solution"​ = chi_"benzene"​P_"benzene"^circ​ + chi_"toluene​"P_"toluene"^circ`

`P_"solution"`​ = (0.541)(150) + (0.459)(50)

= 81.15 + 22.95

= 104.1 mm Hg