Calculate the vapour pressure of a solution of 34.0 g of glycerol (C3H8O3) in 500.0 g of water at

Question

25ºC. The vapour pressure of water at 25ºC is 23.76 torr. Assume ideal behaviour.

Answer 1

First, we need to calculate the mole fraction of glycerol in the solution:

Molar mass of glycerol = 3(12.01) + 8(1.01) + 3(16.00) = 92.09 g/mol

Number of moles of glycerol = 34.0 g / 92.09 g/mol = 0.3695 mol

Molar mass of water = 2(1.01) + 16.00 = 18.02 g/mol

Number of moles of water = 500.0 g / 18.02 g/mol = 27.75 mol

Mole fraction of glycerol = 0.3695 mol / (0.3695 mol + 27.75 mol) = 0.0132

Now we can use Raoult's law to calculate the vapour pressure of the solution:

P = X_water * P_water

where P is the vapour pressure of the solution, X_water is the mole fraction of water in the solution, and P_water is the vapour pressure of pure water.

Mole fraction of water = 1 - 0.0132 = 0.9868

P = 0.9868 * 23.76 torr = 23.45 torr

Therefore, the vapour pressure of the solution is 23.45 torr.