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Question

# The total vapour pressure of a 4 mole % solution of ammonia in water at 293 K is 50.0 torr; the vapour pressure of pure water is 17.0 torr at this temperature. Applying Henry's and Raoult's laws, the total vapour pressure for a 5 mole % solution is (in torr).

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Solution

## The given data is Powater=17.0 torr; Ptotal(4 mole % solution) = PNH3+Pwater =50.0 torr xNH3 = 0.04 and ;xwater= 0.96 Now according to Raoult's law; Pwater=xwater P0water = 0.96 ×17.0 torr = 16.32 torr, Pammonia=33.68 torr Now Henry's law constant for ammonia is KH(NH3)=PNH3xNH3 =842 torr Hence, for 5 mole % solution, we have PNH3 =KH (NH3) xNH3 = (842 torr) (0.05) = 42.1 torr Pwater=xwater P0water = (17 torr) (0.95) =16.15 torr Thus, Ptotal (5 mole % solution) = PNH3+ Pwater = 42.1 + 16.15= 58.25 torr

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