Question
At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of pure toluene = 290 torr). The mole fraction of benzene in the vapor above the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the solution.
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PA=XAP'A
0.213
Since the benzene comes out better you will need more Toluene in the solution
to get its mol fraction above the solution
.59 = Pb/[(Pb)(Pt)}
Assumming 1 atm external pressure
.59 = (1-x)745/(1-x)745 + x(290)
=0.641
to get its mol fraction above the solution
.59 = Pb/[(Pb)(Pt)}
Assumming 1 atm external pressure
.59 = (1-x)745/(1-x)745 + x(290)
=0.641
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