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A mixture of 0.100 mol of N2 and 0.200 mol of O2 is collected over H2O at an...

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lak-86 | Student, Undergraduate | Salutatorian

Posted August 21, 2013 at 4:54 AM via web

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A mixture of 0.100 mol of N2 and 0.200 mol of O2 is collected over H2O at an atmospheric pressure of 750 mm Hg and a temperature of 22 ˚C. What is the partial pressure (in mmHg) of O2 in this mixture?

Vapour pressure of H2O at 22C = 22mmHg

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 21, 2013 at 5:02 AM (Answer #1)

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Since the gasses are collected at atmospheric pressure the total pressure of the gaseous phase is 750mmHg.

At this temperature H2O exerts a pressure of 22mmHg. So the rest is exerted by `O_2 ` and `N_2` .

Pressure by `O_2` and `= 750-22 = 728mmHg`

Amount of `O_2` moles in the mix `= 0.2`

Amount of `N_2 ` moles in the mix  `= 0.1`

Assume that `N_2` and `O_2 ` does not react with each other at the given conditions.

Partial pressure = mole fraction*Pressure

`P_(O_2) = (0.2)/(0.1+0.2)xx728`

`P_(O_2)= 485.33mmHg`

So the answer is 485.33mmHg.

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