The vapor pressure of pure water at 25 °C is 23.8 torr. Determine the vapor pressure (torr) of water
at 25 °C above a solution prepared by dissolving 35 g of urea (a nonvolatile, non-electrolyte, MW =
60.0 g/mol) in 75 g of water.
where have i made a mistake?
35g/60=0.583
75g/18.02=4.162
0.583+4.162=4.745
4.162/4.745=0.88
but my practice test says the answer is 21 where do i go wrong?
please help...
You stopped too soon.
0.88 is the mole fraction of water. Multiply that by the normal vapor pressure of water (23.8) to obtain 20.87 which rounds to 21 mm.
oops, thanks for the help!
To calculate the vapor pressure above the solution of water and urea, you need to use Raoult's law. According to Raoult's law, the vapor pressure of a solvent above a solution is equal to the mole fraction of the solvent multiplied by its pure vapor pressure.
Moles of Urea (35 g):
Moles = Mass / Molar mass = 35 g / 60 g/mol = 0.583 mol
Moles of Water (75 g):
Moles = Mass / Molar mass = 75 g / 18.02 g/mol = 4.162 mol
Total moles in the solution:
Total moles = Moles of Urea + Moles of Water = 0.583 mol + 4.162 mol = 4.745 mol
Mole fraction of Water:
Mole fraction of Water = Moles of Water / Total moles = 4.162 mol / 4.745 mol = 0.875
Vapor pressure above the solution:
Vapor pressure = Mole fraction of Water * Pure vapor pressure of water
Vapor pressure = 0.875 * 23.8 torr = 20.825 torr
Therefore, the vapor pressure of water above the solution is approximately 20.825 torr.
Based on the calculation, it seems like you made a slight error in calculating the mole fraction of water. The correct mole fraction is 0.875, not 0.88. This slight difference in the mole fraction caused the deviation in the final result.
Therefore, the correct answer is approximately 20.825 torr, not 21 torr.
To determine the vapor pressure of water above a solution, we can use Raoult's law, which states that the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent in the solution.
To calculate the mole fraction of water in the solution, we need to find the moles of water and the moles of urea.
Moles of water = mass of water / molar mass of water
Moles of water = 75 g / 18.02 g/mol
Moles of water = 4.162 mol
Moles of urea = mass of urea / molar mass of urea
Moles of urea = 35 g / 60.0 g/mol
Moles of urea = 0.583 mol
Next, we can calculate the mole fraction of water in the solution:
Mole fraction of water = moles of water / total moles
Mole fraction of water = 4.162 mol / (4.162 mol + 0.583 mol)
Mole fraction of water = 0.877
Now we can use Raoult's law to calculate the vapor pressure of water above the solution:
Vapor pressure of water above the solution = mole fraction of water * vapor pressure of water
Given that the vapor pressure of pure water at 25 °C is 23.8 torr:
Vapor pressure of water above the solution = 0.877 * 23.8 torr
Vapor pressure of water above the solution = 20.86 torr
So, based on the calculations, the vapor pressure of water above the solution should be approximately 20.86 torr, not 21 torr.
Therefore, it seems that you have not made an error in your calculations. There might be a discrepancy between the expected answer and the answer given on the practice test. It's always important to double-check the provided answer and, if needed, consult with your instructor for clarification.