Question

The vapor pressure of pure water at 25C is 23.77 mm Hg. Calculate the change in vapor pressure when 60 g glucose (MW = 180 g/mole) is added to 500 mL of water.

Answer 1

According to Raoult's Law

P = X solvent X P°Solv

Where X is the mole fraction of solvent and P° vapor pressure of pure solvent

Given; vapor pressure of pure water at 25C is 23.77 mm Hg

60 g glucose (MW = 180 g/mole) is added to 500 mL of water.

Number of moles of glucose: 60/180 = 0.33

Number of moles of water: 500 mL = 500 g (because density is 1g/cc = 500/18(MW = 18 g/mole) = 27.8

Mole fraction of water = 27.8/ (27.8+0.33) = 0.989

P = 0.989 X 23.77 = 23.51 mm Hg