The vapor pressure of water at 90°C is 0.692 atm.What is the vapor pressure (in atm)of a solution made by dissolving 1.28 mole(s)of CsF(s)in 1.00 kg of water? Assume that Raoult's law applies.
A) 0.661 atm
B) 0.676 atm
C) 0.692 atm
D) 0.724 atm
E) none of these

Question

Quizplus · Accepted Answer

Using Raoult's law:

Psolution = Xsolvent * P°solvent

where Psolution is the vapor pressure of the solution, Xsolvent is the mole fraction of water (solvent), and P°solvent is the vapor pressure of pure water at the same temperature.

Since CsF is a non-volatile solute, it does not contribute to the vapor pressure of the solution. Therefore, we only need to consider the water component.

Molar mass of H2O = 18 g/mol
Mass of 1 kg of water = 1000 g
Number of moles of water = 1000 g / 18 g/mol = 55.56 mol

Mole fraction of water:
Xwater = moles of water / total moles of solution
Xwater = 55.56 mol / (55.56 mol + 1.28 mol) = 0.977

Using the vapor pressure data given:

Psolution = Xwater * P°water
Psolution = 0.977 * 0.692 atm = 0.676 atm

Therefore, the answer is A) 0.661 atm.

The Vapor Pressure of Water at 90°C Is 0