Calculate the vapor pressure of the resulting ethylene glycol solution.

Question

107.1 g of non-volatile ethylene glycol (62.07 g mol ) is added to 336 g of water at room temperature. The vapor pressure of water at room temperature is 23.76 Torr. Calculate the vapor pressure of the resulting ethylene glycol solution.

Thank you!

Answer 1

Raoult's Law

P = XPº

P = vapor pressure of solution = ?

X = mole fraction of the solvent

Pº = vapor pressure of the pure solvent

So, first we'll find the mole fraction of water in the solution:

moles ethylene glycol (EG) = 107.1 g x 1 mol EG/62.07 g = 1.725 moles EG

moles H₂O = 336 g x 1 mol / 18 g = 18.7 moles H2O

Total moles = 1.725 + 18.7 = 20.4 moles

Mole fraction H₂O = X = 18.7 / 20.4 = 0.917

P = vapor pressure of solution = XPº = (0.917)(23.76 torr)

P = 21.8 torr (3 sig. figs.)