Question

The vapor pressure of pure water at 250C is 23.77 torr. What is the vapor pressure of water above a solution that is 1.500 m glucose, C6H12O6?

Answer 1

Answer:

Vapor pressure of water = 23.14torr

Explanation:

When you made a solution, vapor pressure decreases following Raoult's law:

$P_{solution} = X_{solvent} P_{solvent}$

Where P is vapor pressure and X mole fraction

As vapor pressure of water is 23.77torr we must find the mole fraction of water knowing the solution is 1.500m glucose (That is 1.500 moles of glucose per kg of water = 1000g of water).

1000g of H₂O are, in moles (Molar mass: 18.02g/mol):

1000g H₂O ₓ (1mole / 18.02g) = 55.5 moles of H₂O.

As we know now the solution contains 55.5 moles of water and 1.5 moles of glucose. Thus, mole fraction of water (Solvent) is:

$X_{H_2O} = \frac{55.5molesH_2O}{55.5molesH_2O + 1.5 molesGlucose} = 0.9737$

Replacing in Raoult's law, pressure of water above the solution is:

$P_{solution} = X_{solvent} P_{solvent}$

$P_{solution} = 0.9737*23.77torr$

Vapor pressure of water = 23.14torr