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The vapour pressure of water at 27 deg C is 0.24 atm. For the equilibrium H2O(l) <=> H2O(g) at 27 deg C, calculate Kp (in atm) and Kc (in M). (R = 0.08 L-atm/K-mol)

1. Kp = 0.24 atm, Kc = 0.010 M
2. Kp = 0.24 atm, Kc = 0.24 M
3. Kp = 0.08 atm, Kc = 0.24 M
4. Kp = 0.24 atm, Kc = 0.08 M

Correct answer: Kp = 0.24 atm, Kc = 0.010 M

Solution

Kp = p(H2O gas) = 0.24 atm. Then Kc = Kp/(RT) = 0.24/(0.08*300) = 0.24/24 = 0.010 M.

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