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 17.4 g of biphenyl in 26.0 g of benzene?

Answer 1

first we have to calculate the mole fraction of benzene after adding the biphenyl

no of moles of biphenyles = 17.4 g /154.2 g/mol = 0.113 moles C12H10

no of molea of benzene = 26.0 g / 78.1 g/mol= 0.333moles C6H6

Total moles = moles C12H10 + moles C6H6 = 0.113 + 0.333 = 0.446 moles

mole fraction C6H6 = (moles C6H6 / total moles) = 0.333 / 0.446 = 0.747

this means that the vapor pressure will be 74.7% of the vapor pressure of pure benzene since the solution contains only 74.7 mole % benzene. The biphenyl contributes no vapor pressure (nonvolatile).

P = (x C6H6)(Po C6H6) where x C6H6 is the mole fraction of C6H6 and Po C6H6 is the vapor pressure of pure C6H6.

P = (0.747)(100.84 torr) = 75.3 torr.