A solution is made by mixing 43.0 mL of ethanol, C2H6O, and 57.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 ∘C?

Question

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Answer 1

We need to know the vapor pressure of pure ethanol (EtOH) and pure water (H2O). Then you can use the mole fractions of the two ingredients to find the vapor pressure of the mixture. Also, to find the mole fraction of ethanol, we'd need to know the density of EtOH. We can probably assume the density of water to be 1 g /ml.

Looking up vapor pressures and densities at 25ºC, I find the follow:

vapor pressure pure EtOH = 0.0790 atm

density of pure EtOH= 0.785 g / ml

vapor pressure pure water = 0.0313 atm

density of pure water = 1 g / ml

moles EtOH = 43.0 ml x 0.785 g / ml x 1 mol / 46 g = 0.0734 moles

moles water = 57.0 ml x 1 g / ml x 1 mol / 18 g = 3.17 moles

Total moles = 3.24 moles

Mole fraction EtOH = 0.0734 mol / 3.24 mol = 0.0227

Mole fraction H2O = 3.17 mol / 3.24 = 0.978

Vapor pressure of mixture = P_T = X_EtOH x Pº_EtOH + X_H2O + Pº_H2O

P_T = (0.0227 * 0.0790 atm) + (0.987 * 0.0313 atm) = 0.00179 atm + 0.0309 atm

P_T = 0.0327 atm