What is the volume of 2.0 moles of oxygen collected over water at 23 degrees Celsius and 798 torr? (at 23 degrees Celsius, the vapor pressure of water is 0.0277 atm or 21.1 torr)

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jerichorayel's profile pic

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We can solve this problem using the ideal gas equation. It is expressed as:

`PV = nRT`

First we have to correct the pressure that is exerted by the gas since we are collecting it in water. 

`P_(t o t a l) = P_(O_2) + P_(H_2O)`

`P_(O_2) =P_(t o t a l) -P_(H_2O)`

`P_(O_2) = 798 - 21.1`

`P_(O_2) = 776.9 t o r r`

 

Given:

`P = (776.9 t o r r)/(760) = 1.022 atm`

`n = 2.0 mol es`

`R = 0.08206 (atm-L)/(mol-K)`

`T = 23 + 273.15 = 296.15 K`

 

`V = ?`

 

`PV = nRT`

`V = (nRT)/(P)`

`V = (2.0*0.08206*296.15)/(1.022)`

`V = 47.6La n s w e r`

 

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