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The vapour pressure of water at 90 oC is 0.692 atm. What is the vapour pressure (in atm) of a solution made by dissolving 1.00

The vapour pressure of water at 90 oC is 0.692 atm. What is the vapour pressure (in atm) of a solution made by dissolving 1.00 mole of CsF(s) in 1.00 kg of water? Assume Raoult's Law applies.

0.688

0.680

0.692

0.656

0.668

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