posted by Daiyo .
The normal boiling point of ethanol is 78.4°C. When 9.16 g of a soluble nonelectrolyte was dissolved in 100. g of ethanol, the vapor pressure of the solution at that temperature was 7.40 102 Torr.
(a) What are the mole fractions of ethanol and solute?
(b) What is the molar mass of the solute?
The vapor pressure is actually 7.40 x 10^2
The vapor pressure of ethanol (any liquid at its boiling point) is 760 torr. So
Psoln = Xethanol*Poethanol.
Xethanol = P/Po = 740/760 = ??
Xethanol + Xsolute = 1
(b). You know 100 g ethanol. Change to moles ethanol.
nethanol/(nethanol+nsolute) = Xethanol (from a part).
Solve for nsolute(moles solute).
Then moles = grams/molar mass. Solve for molar mass.
Post your work if you get stuck.
Thanks for the help DrBob
I forgot the pressure of any liquid at boiling point