Determine the vapor pressure of a solution at 25°C that contains 76.6 g of glucose (C6H12O6) in 250.0 mL of water (Density of water = 1.0g/mL). The vapor pressure of pure water at 25°C is 23.8 torr

Answer 1

The vapor pressure of a solution will be lower than the vapor pressure of the pure solvent. It will be equal to the vapor pressure of the pure solvent times the mole fraction of that solvent.

moles glucose = 76.6 g glucose x 1 mol / 180 g = 0.4256 moles

moles water = 250 .0 ml x 1 g / ml x 1 mol / 18 g = 13.89 moles

Total moles = 14.32 moles

Mole fraction of water = 13.89 mols / 14.32 mols = 0.97

Vapor pressure of solution = 0.97 x 23.8 torr

Vapor pressure of solution = 23.1 torr