Question

A student made a solution by dissolving 0.144g of an unknown in 10.00g of benzene, C6H6, and found the vapor pressure solution to be 94.35mm Hg. The vapor pressure of pure benzene is 95.00mm Hg at the temperature of the experiment. (Show your work and use the factor labor method)

a. what is the mole fraction of benzene in the solution?

b. how many moles of benzene were in the solution?

c. how many moles of unknown were in the solution?

d. calculate the molar mass of the unknown.

Answer 1

= = Unkonwn Benzene Here Solute Benzene According to Raoult's law Vapor pressure of sol (P sol) Psol Vapor Pressure of pure Benzene (Po) Vapor Pressure of Solution (P sol) Mole fraction of Benzene (Xo) Psol Xo = = = = (Mole fraction of solvent) * (vapor pressure of pure solvent Xo * Po 95.00 mmhg 94.35 mmhg = = Xo* P soln РО 94.350 95.00 Mole fraction of Benzene Xo = torr * torr = 0.993 Mass of Benzene Molar mass of Benzene Moles of Benzene 10.00 g 78.11 g mol-1 Mass of solute Molar mass of solute 10.00 78.11 0.128 g g mol-1 mol Benzene = Mole fraction of Benzene (x Benzene) = n solute n Benzene 0.1280 n solvent n solute +n benzene n solvent x Benzene 0.1280 - 0.993 0.000882 mol Mass of solute Moles of Solute 0.1440 g 0.0008820 mol 163.3 g/mol (Moles of Unknown) Molar Mass of solute = =

a. what is the mole fraction of benzene in the solution = 0.993

b. how many moles of benzene were in the solution = 0.128 mol

c. how many moles of unknown were in the solution = 0.000882 mol

d. calculate the molar mass of the unknown = 163.3 g/mol