Question

Benzene (C₆H₆) and toluene (C₆H₅CH₃) form an ideal solution. At 313 K, the vapour pressures of pure benzene and toluene are 160 mm Hg and 60 mm Hg respectively. Calculate the partial pressures of benzene and toluene and the total pressure of the solution when equal masses of benzene and toluene are mixed at 313 K.

Answer 1

Given: Benzene (C₆H₆) and toluene (C₆H₅CH₃) form an ideal solution.

Temperature = 313 K

Vapour pressure of pure benzene `P_"benzene"^circ` ​= 160 mm Hg

Vapour pressure of pure toluene `P_"benzene"^circ` = 60 mm Hg

Equal masses of benzene and toluene,

Molar mass of benzene = 78 g/mol

Molar mass of toluene = 92 g/mol

Let us assume 78 g of each is taken.

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_"benzene"​ = chi_"benzene"​P_"benzene"^circ`​

= 0.541 × 160

= 86.56 mm Hg

`P_"toluene​" = chi_"toluene​"P_"toluene"^circ`

= 0.459 × 60

= 27.54 mm Hg

`P_"total"​ = P_"benzene" + P_"toluene​"`

= 86.56 + 27.54

= 114.1 mm Hg