At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution.
(a) What is the composition in mole fractions of a solution that has a vapor pressure of 40. torr at 20°C?
(b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
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Using Algebra!
(a) What is the composition in mole fractions of a solution that has a vapor pressure of 40. torr at 20°C?
Let X be the mole fraction of benzene in this ideal solution. Thus 1-X is the mole fraction of toluene in this ideal solution.
The following is given:
75X + 22(1-X) = 40, or:
53X = 18, and X = 18/53.
Benzene's mole fraction is 18/53, and toluene's mole fraction is 35/53.
(b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
(75*18/53)/40 = 63.7%
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Vapor Pressure Mole Fraction
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Hi bill! Your attempt is not correct. Convert from Torr to atm: 526/760 = 0.692 atm p solution = p water *Xwater p solution is the vapor pressure of the component in the mixture. p water is the vapor pressure of the pure component. Xwater is the molar fraction of the component in the mixture. p solution = 0.660 atm at 90 °C p water = 0.692 atm at 90 °C 0.660 = 0.692* Xwater Xwater = 0.660/0.692 = 0.954 Xwater + Xglucose = 1 Xglucose = 1 - Xwater Xglucose = 1 – 0.954 = 0.046 Answer: mol fraction of glucose = 0.046 at 90 °C Bye from Italy, C6H6 Signature: Asking is legitimate and choosing the Best Answer is good manners. Thank you.
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