If the vapor pressure of pure benzene is 24.4 kPa at 40.0 °C, then what will the vapor pressure of the solution be in kPa?

Question

0.0711 kg of biphenyl (C₁₂H₁₀) is dissolve in benzene (C₆H₆) to create a solution with a total volume of 350.0 mL. (Assume the change in volume is negligible) If the vapor pressure of pure benzene is 24.4 kPa at 40.0 °C, then what will the vapor pressure of the solution be in kPa? (Consider biphenyl to be nonvolatile and the density of benzene is 0.877 g/mL)

Answer 1

When you add a non volatile solute (biphenyl) to a solvent (benzene), the vapor pressure of the benzene is reduced. It is reduced by an amount equal to the mole fraction of the solute that is present. So, what we need to do is find the moles of solute (biphenyl), the moles of solvent (benzene), and then the total moles. Then we'll find mole fractions.

moles biphenyl = 0.0711 kg x 1000 g / kg x 1 mol biphenyl / 154.2 g = 0.461 mols biphenyl

moles benzene = 350.0 ml x 0.877 g / ml x 1 mol benzene / 78.11 g = 3.93 mols benzene

Total mols = 0.461 + 3.93 = 4.39 mols

Mole fraction of benzene = 3.93 mols / 4.39 mols = 0.895

Vapor pressure of the solution = vapor pressure of pure benzene x mole fraction of benzene

Vapor pressure of the solution = 24.4 kPa x 0.895 = 21.8 kPa