12. 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 3.68 mole(s) of CsF(s) in 1.00 kg of water? Assume that Raoult's law applies.

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of = 1.00 kg = 1000 g

Moles of = 3.68 mole

Molar mass of = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of .

Now we have to calculate the mole fraction of

Now we have to partial pressure of solution.

According to the Raoult's law,

where,

= vapor pressure of solution

= vapor pressure of water = 0.692 atm

= mole fraction of water = 0.938

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

43.452

step-by-step explanation:

284 times .153 = 43.452