Question

Benzene and toluene form nearly ideal solutions. Consider an equimolar solution of benzene and toluene. At 20°C the vapour pressures of pure benzene and toluene are9.9 kPa and 2.9 kPa, respectively. The solution is boiled by reducing the external pressure below the vapour pressure.
Calculate:
(a) The pressure when boiling begins,
(b) The composition of each component in the vapour, and
(c) The vapour pressure when only a few drops of liquid remain

Answer 1

(a) By Raoult's Law (P_{tot}= x_{B}P_{B}^{*} + x_{T}P_{T}^{*})

Ptot = .5(9.9kPa) + .5(2.9kPa) = 6.4 kPa

Boiling begins when the external pressure equals the vapor pressure

(b) By Dalton's Law (y_{B} = (x_{B}P_{B}^{*})/P_{tot})

yB = (.5*9.9)/6.4 = .77 Benzene in vapor .23 Toluene

(c) When almost all your solution has been converted into vapor, the vapor is now .50 Benzene and .50 Toluene, (i.e. if you start with equimolar liquidyou end with equimolar gas).

(.5 = (2.9x_{T})/P_{tot}) (.5=(9.9x_{B})/P_{tot})

re-arrange

(P_{tot} = 5.8x_{T}) &&(P_{tot} = 19.8x_{B})

(5.8x_{T} = 19.8x_{B})

(5.8(1-x_{B}) = 19.8x_{B})

solve for xB = .23 Benzene in the liquid and .77 Toluene

Calculate your vapor pressure from Raoult's law

P = .23(9.9 kPa) + .77(2.9 kPa) = 4.51 kPa