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 19.4 g of biphenyl in 30.3 g of benzene?

Answer 1

The solution is as given below,

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

19.4 g C12H10 x (1 mole C12H10 / 154.2 g C12H10) = 0.126 moles C12H10

30.3 g C6H6 x (1 mole C6H6 / 78.1 g C6H6) = 0.388 moles C6H6

Total moles = moles C12H10 + moles C6H6 = 0.126 + 0.388 = 0.514 moles

mole fraction C6H6 = (moles C6H6 / total moles) = 0.388 / 0.514 = 0.755

What that means is that the vapor pressure will be 75.5% of the vapor pressure of pure benzene since the solution contains only 75.5 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 = (0.755)(100.84 torr) = 76.13 torr.

Thus the vapor pressure of solution is 76.13 torr.