Question
Calculate the vapour pressure of a solution of 34.0 g of glycerol (C3H8O3) in 500.0 g of water at
25ºC. The vapour pressure of water at 25ºC is 23.76 torr. Assume ideal behaviour.
25ºC. The vapour pressure of water at 25ºC is 23.76 torr. Assume ideal behaviour.
Answers
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.
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.
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