Question

Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 11.3 g of biphenyl in 25.9 g of benzene?

Answer 1

Number of Moles of Biphenyl = (11.3 g ) / (154.21 g/mol) = 0.0734 moles
Number of Moles of C6H6 = (25.9 g ) / (78.11 g/mol) = 0.33158 moles

Supposing the mixture in question to be at 25 °C:

Vapor Pressure of solution = (Vapor pressure of Benzene * Moles of Benzene) / (Moles of solute + Moles of solvent)
(100.84 torr) x (0.34567 mols C6H6) / (0.33158 mols + 0.0734 mols) = 86.07 torr

Vapor Pressure of solution = 86.07 torr