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 15.4 g of biphenyl in 29.7 g of benzene?

Answer 1

first we have to claculate the mole fraction of benzene after adding the biphenyl (molar mass = 154.2 g/mole).

moles of C12H10= 17.9 g/154.2g = 0.116 moles C12H10

moles of C6H6 =27.4 g/ 78.1 g = 0.351 moles C6H6

Total moles 0.116 + 0.351 = 0.467

mole fraction C6H6 = (moles C6H6 / total moles) = 0.351 / 0.467 = 0.752

that means the vapor pressure will be 75.2% of the vapor pressure of pure benzene since the solution contains only 75.2 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 = (0.752)(100.84 torr) = 75.8 torr.