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11 March, 14:44

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.

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  1. 11 March, 15:12
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    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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