Two kilograms of a mixture of 40-60 (by weight) benzene and toluene is at 60C. A

Question

Two kilograms of a mixture of 40-60 (by weight) benzene and toluene is at 60C. As the total pressure on the system is reduced:
At what pressure (mmHg) will the first bubble of vapor appear?
What will be the molar composition of this first bubble of vapor? Two kilograms of a mixture of 40-60 (by weight) benzene and toluene is at 60C. As the total pressure on the system is reduced:
At what pressure (mmHg) will the first bubble of vapor appear?
What will be the molar composition of this first bubble of vapor?
At what pressure (mmHg) will the first bubble of vapor appear?
What will be the molar composition of this first bubble of vapor?

Explanation / Answer

Solution:

Using Antonie equation pure-component vapor pressures for benzene and toluene at 60oC. We obtain:

P*benzene = 9080.432 mm Hg          P*toluene= 833.46 mm Hg

Molecular wt of benzene=78.11

Molecular wt of toluene=92.14

Moles of benzene=2000*0.4/78.11 = 10.25 gm moles

Moles of toluene=2000*0.6/92.14 = 13.02

xbenzene = 10.25/(10.25+13.02)=0.44

xtoluene = 13.02/(10.25+13.02)=0.56

Total pressure by Raoult's Law: This is the pressure at which first bubble of vapor appear.

       P = xbenzeneP*benzene + xtolueneP*toluene = xbenzene(P*benzene - P*toluene) + P*toluen   = (0.44)(9080.42- 833.46) + 833.46 = 4462.12 mm Hg

vapor composition:

Ybenzene = xbenzeneP*benzene(T)/P = (0.44)(9080.432)/4462.12) = 0.895

Ytoluene= 1-0.895=0.105

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