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 1.17 mole (s) of csf (s) in 1.00 kg of water? assume that raoult's law applies.

We shall proceed by finding the number of moles of H2O. Now 1 kg of water = 1000 g = 1000/18 = 55.55 mol The number of moles of CsF = 1.17 mol Hence the mole fraction of water in the solution = 55.55 / (55.55 + 1.17) = 0.979 From Raoult's Law, It follows that the vapour pressure of water = 0.979 x 0.692 = 0.6771 atm