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How do I solve this Chem problem? Partial vapor pressures of solutions?
Toluene and Benzene make ideal solutions, What is the overall vapor pressure lowering when benzene is mixed with toluene to make a .7 mole fraction of benzene (with .3 mole fraction for toluene). The vapor pressure for benzene is 75mmHg and the vapor pressure for toluene is 22mmHg. What is the mole fraction of benzene in the vapor?
Please help in whatever way you can. This is one of the practice problems for studying for exams and I have no idea where to start.
1 Answer
- HPVLv 78 years agoFavorite Answer
I am assuming that you are adding toluene to the benzene, and want to know what the vapor pressure lowering of pure benzene is. (Note: x = mole fraction).
P due to benzene = (x benzene)(P pure benzene) = (0.7)(75 mmHg) = 53 mmHg
P due to toluene = (x toluene)(P pure toluene) = (0.3)(22 mmHg) = 7 mm Hg
Total vapor pressure = 53 mmHg + 7 mmHg = 60 mmHg . . .that's 15 mmHg lower than for pure benzene.
x benzene in vapor = (P due to benzene / total P) = (53 mm Hg / 60 mmHg) = 0.88
