Benzene and toluene form nearly ideal solutions. Consider an equimolar solution

Question

Benzene and toluene form nearly ideal solutions. Consider an equimolar solution of benzene and toluene. At 20 degree C the vapor pressure of pure benzene and toluene are 9.9 kPa and 2.9 kPa, respectively. The solution is boiled be reducing the external pressure below the vapor pressure. Calculate the pressure when the boiling begins, calculate the mole fraction of each component in the vapor at this point, Calculate the vapor pressure when only a few drops of liquid remain. Assume that the rate of vaporization is slow enough so that the temperature remains at 20 degree C.

Explanation / Answer

let moles of benzene= moles of toluene =1

mole fractions : Benzene = 1/2 =0.5= toluene

from raoults law , y1P= x1p1sat and y2P= x2p2sat where y1, y2 are mole fraction in the vapor phase , x1, x2 are mole fractions in the liquid phase.

P= x1p1sat + x2p2sat = 0.5*9.9+0.5*2.9=6.4 Kpa, pressure at which boiling begins

from y1p= x1p1sat, y1= 0.5*9.9/6.4= 0.77 and y2 = 1-0.77=0.23

since only few drops of liquid remain, the pressure is 9.9 Kpa

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