Help me solve this problem??

Question

You have prepared a solution by dissolving 16.323g of dextrose (C6H12O6) in 23.50mL 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 0.0317atm.

Answer 1

The vapor pressure of the solvent (water) will decrease when you add the solute (dextrose). The amount of this decrease will depend on the concentration of the dextrose. You can review the theory behind this by reading about Raoult's Law. This states that the the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature scaled by the mole fraction of the solvent present. Or put another way, we will find the mol fraction of water in the solution and then determine the vapor pressure.

molar mass dextrose = 180 g/mol

moles dextrose present = 16.32 g x 1 mol / 180 g = 0.0907 mols

molar mass H2O = 18 g/ mol

moles H2O present = 23.50 ml x 1 g / ml x 1 mol 18 g = 1.306 mols (assuming a density of 1)

TOTAL MOLES = 0.0907 + 1.306 = 1.397 mols

mol fraction of H2O = 1.306 mol / 1.397 mol = 0.9349

Vapor pressure of solution = 0.9349 x 0.0317 atm = 0.0296 atm