How do you calculate this?

Question

You have prepared a solution by dissolving 14.521g of dodecane (C12H26) in 34.50mL of cyclohexane (C6H12) at 25°C. What is the expected vapor pressure (in units of mmHg) of this solution? At 25°C, the density of cyclohexane is 0.7792g/mL and its vapor pressure is 95.1mmHg. Assume that dodecane is a non-volatile solute. (steps)

Answer 1

When you dissolve a substance in a solvent, the vapor pressure of the pure solvent will be reduced. See Raoult's law.

The decrease in vapor pressure is related to the mol fraction of the solvent. In a pure solvent it is 1.00, of course. We can find the mol fraction of cyclohexane in the present problem by first finding mols of cyclohexane, then the mols of dodecane. Add them to get total mols and then divided mols cyclohexane by the total mols to get the mol fraction of cyclohexane.

mols cyclohexane (molar mass = 84.16 g/mol):

34.50 ml C₆H₁₂ x 0.7792 g / ml x 1 mol / 84.16 g = 0.3194 mols

mols dodecane (molar mass = 170.34 g / mol):

14.521 g C₁₂H₂₆ x 1 mol / 170.34 g = 0.085247 mols

Total mols = 0.3194 mols + 0.085247 mols = 0.4046 mols

Mole fraction of cyclohexane = 0.3194 mol / 0.4046 mol = 0.7894

Now, according to Raoult's law, the vapor pressure of the solution will be the vapor pressure of the pure solvent (95.1 mm Hg) times the mole fraction of the pure solvent (0.7894)

P_soln = X_C6H12 x Pº_C6H12

P_soln = (0.7894)(95.1 mm Hg) = 75.1 mm Hg (3 sig, figs.)