Question

The vapor pressure of pure water at 25 °c is 23.8 torr. determine the vapor pressure (torr) of water at 25 ºc above a solution prepared by dissolving 25 g of urea (a nonvolatile, non-electrolyte, mw = 60.0 g/mol) in 75 g of water. pregunta 11trieu-ne una:

a. 22
b. 0.42
c. 27
d. 2.9
e. 0.91

Answer 1

To find the vapor pressure of water at 25 ºC above a solution with dissolved urea, calculate the moles of each component, find the mole fraction of water, and apply Raoult's Law. The resulting vapor pressure will be lower than 23.8 torr, the vapor pressure of pure water at 25 ºC.

Step-by-step explanation:

To determine the vapor pressure (torr) of water at 25 ºC above a solution created by dissolving 25 g of urea, we use Raoult's Law. According to Raoult's Law, the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution.

First, calculate the moles of urea by using its molar mass (60.0 g/mol):
moles of urea = 25 g / 60.0 g/mol = 0.4167 moles

Next, calculate the moles of water using its molar mass (18.0 g/mol):
moles of water = 75 g / 18.0 g/mol = 4.1667 moles

Then, find the mole fraction of water (Xwater):
Xwater = moles of water / (moles of water + moles of urea) = 4.1667 / (4.1667 + 0.4167)

Finally, apply Raoult's Law:
vapor pressure of solution = vapor pressure of pure water at 25 ºC × Xwater
vapor pressure of solution = 23.8 torr × (4.1667 / (4.1667 + 0.4167))

By calculating this, we find that the vapor pressure of the solution is slightly lower than the vapor pressure of pure water at 25 ºC, which is 23.8 torr.