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 10.6 g of biphenyl in 33.1 g of benzene?

Answer 1

We need to calculate the mole fraction of benzene after adding the biphenyl (molar mass = 154.2 g/mole).

10.6 g C12H10 x (1 mole C12H10 / 154.2 g C12H10) = 0.0687 moles C12H10

33.1 g C6H6 x (1 mole C6H6 / 78.1 g C6H6) = 0.4238 moles C6H6

Total moles = moles C12H10 + moles C6H6 = 0.0687 + 0.4238 = 0.4925

mole fraction C6H6 = (moles C6H6 / total moles) = 0.4238 / 0.4925 =0.8605

What that means is that the vapor pressure will be 86.05% of the vapor pressure of pure benzene since the solution contains only 86.05 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 = (0.8605)(100.84 torr) = 86.77 torr.