80.0 L of oxygen is collected over water at 30.0 °C. The atmospheric pressure in the room is 96.00 kPa. (a) What is the total pressure of this mixture? [ans: 96.00 kPa] (b) What is the partial pressure of the oxygen gas? [ans: 91.76 kPa]
The answers to the questions are posted beside the question. I don't know how to get to those answers.
The answers to the questions are posted beside the question. I don't know how to get to those answers.
1 Answer
See the explanation below...
Explanation:
The first answer is just a matter of understanding that if the surrounding atmospheric pressure is 96 kPa, then the total pressure of the collected sample must be the same, as the collection apparatus would have been open to the air in some way.
Secondly, since the oxygen was collected over water, it will actually consist of a mixture of
Likely, you are expected to determine the vapour pressure of the water by looking it up.
A site like this one will give you this value:
http://www.msduncanchem.com/Reference_Tables/water_vapor_pressure_chart.htm
At 30.0 °C, the vapour pressure of water is 4.25 kPa
To find the vapour pressure of the oxygen along, you subtract the pressure of the water from the total pressure:
96.0 - 4.25 = 91.75 kPa
