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A sample of damp air in a 1.00 L container exerts a pressure of 762.0 torr...

A sample of damp air in a 1.00 L container exerts a pressure of 762.0 torr at 48.5 °C; when it is cooled to -10.0 °C, the pressure falls to 607.1 torr as the water condenses and freezes. What mass of water was present? Assume ideal gas behavior and that the vapor pressure of ice at -10.0 °C is negligible

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Answer #1

The key point here is that ice, H20 (s) exerts almost no vapor pressure

P tot = Pair + PH2O(g)   initially

762 = (607.1)*(321.5/263) + PH2O(g)

PH2O(g) = 762 - (607.1)*(321.5/263)

= 19.86 torr

Mass = (19.86/(760*0.0821*293))*18

= 0.01955 g

= 19.55 mg

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