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

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