Determine the vapor pressure of this solution at 25˚C, given that the vapor pressure of pure water at this temperature is 23.8 mm Hg.

Question

Antifreeze is usually composed of a 1:1 mixture, by weight, of water and ethylene glycol (MW = 62.068 g/mol). Determine the vapor pressure of this solution at 25˚C, given that the vapor pressure of pure water at this temperature is 23.8 mm Hg.

Answer 1

RAOULT'S LAW

The vapor pressure of the solution will vary as the mole fraction of solute and water varies. So, we will determine the mole fraction of H2O and then multiply that by the vapor pressure of pure H2O.

XP_o = P_soln

Since they tell us we have a 1:1 mixture by weight, let us just assume that we have 62.098 g of ethylene glycol (1 mole) and 62.098 g of H2O. We can now determine moles of water.

62.098 g H2O x 1 mol / 18.02 g = 3.45 moles

Total moles = 1.00 mol ethylene glycol + 3.45 moles H2O = 4.45 moles

Mole fraction of H2O = X_H2O = 3.45 mol / 4.45 mol = 0.775

Vapor pressure of solution = 0.775 x 23.8 mm Hg = 18.5 mm Hg