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.
A) 0.88
B) 27
C) 21
D) 3.3
E) 2.9

Question

Quizplus · Accepted Answer

First, we need to calculate the mole fraction of water and urea in the solution.

Moles of solute (urea) = mass/molar mass = 35/60.0 = 0.5833 mol
Moles of solvent (water) = mass/molar mass = 75/18.0 = 4.17 mol

Total moles in solution = moles of solute + moles of solvent = 4.17 + 0.5833 = 4.7533 mol

Mole fraction of water = moles of water / total moles = 4.17 / 4.7533 = 0.8773
Mole fraction of urea = moles of urea / total moles = 0.5833 / 4.7533 = 0.1227

Now we can use Raoult's law to calculate the vapor pressure of the solution.

P solution = X water * P water

where P water is the vapor pressure of pure water and X water is the mole fraction of water in the solution.

P solution = 0.8773 * 23.8 torr = 20.9 torr

Finally, we use the equation for the vapor pressure lowering (ΔP = P° - P solution) to find the vapor pressure of water above the solution.

ΔP = P° - P solution
ΔP = 23.8 torr - 20.9 torr
ΔP = 2.9 torr

Therefore, the vapor pressure (torr) of water at 25 °C above the solution is 2.9 torr. The best answer choice is C.

The Vapor Pressure of Pure Water at 25 °C Is