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 1 mole of benzene and 4 moles of 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

1 mole benzene, 4 moles toluene

Total moles = 1 + 4 = 5 moles

`chi_"benzene" = 1/5`  = 0.2

`chi_"toluene" = 4/5` = 0.8

By using Raoult’s law

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

= 0.2 × 160

= 32 mm Hg

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

= 0.8 × 60

= 48 mm Hg

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

= 32 + 48

= 80 mm Hg