Raoult's Law Calculations

Question

The vapor pressure of water is 23.76 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in water is glucose.

Calculate the vapor pressure of the solution at 25 °C when 14.85 grams of glucose, C₆H₁₂O₆ (180.2 g/mol), are dissolved in 225.8 grams of water.

water = H₂O = 18.02 g/mol.

VP(solution) =  mm Hg

Answer 1

The vapor pressure of the solution will be equal to the mole fraction of water x the vapor pressure of pure water. This is Raoult's Law. So, first, let us find the mole fraction of water in the solution.

mole fraction H₂O = X_H2O

moles Glucose = 14.85 g glucose x 1 mol glucose/180.2 g = 0.08241 moles glucose

moles H₂O = 225.8 g H₂O x 1 mol/18.02 g = 12.53 moles H₂O

Total moles present = 0.08241 + 12.53 = 12.61 moles

X_H2O = 12.53 mol/12.61 mol = 0.9937

Vapor pressure of solution = X_H2O • Pº_H2O where Pº = vapor pressure of pure H2O = 23.76 mm Hg

VP(solution) = 0.9973 • 23.76 mm Hg = 23.61 mm Hg