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Question

You have prepared a solution by dissolving 16.425g of dextrose (C6H12O6) in 21.50mL of water at 25°C. What is the expected vapor pressure (in units of atm) of this solution? The vapor pressure of pure water at 25°C is 0.0313atm.

Answer 1

Molar mass dextrose = 180 g / mol

Molar mass H2O = 18 g / mol

Raoult's law: vapor pressure = mole fraction x vapor pressure of pure substance

So, we need to find the mol fraction of water and then multiply that by 0.0313 atm to get out answer.

The mole fraction of water is mols H2O / total mols

mols H2O = 21.50 mls x 1 g / ml x 1 mol / 18 g = 1.194 mols H2O

mols dextrose = 16.425 g x 1 mol / 180 g = 0.09125 mols dextrose

Total mols = 1.194 + 0.09125 = 1.285 mols

Mole fraction H2O = 1.194 / 1.285 = 0.929

Vapor pressure of solution = 0.929 x 0.0313 atm = 0.0291 atm