Asked by Natalie

A sample of oxygen gas was collected via water displacement. Since the oxygen was collected via water displacement, the sample is saturated with water vapor. If the total pressure of the mixture at 26.4 °C is 725 torr, what is the partial pressure of oxygen? The vapor pressure of water at 26.4 °C is 25.81 mm Hg.
Answered by DrBob222
Ptotal = pO2 + pH2O
Answered by Natalie
I got .988atm, but what do I do with the temp?
Answered by DrBob222
The temperature is used to look in a table of T vs vapor pressure. It tells you the vapor pressure of the water (in this case 25.81 mm Hg) at 26.4 C.
I don't agree with your answer. And why in the world would you change mm and torr to atm if the problem didn't ask for the answer in atm?
Answered by Natalie
Ok then, I got 750.81 torr.

I know that Partial pressure = total pressure x the mole fraction

Does this mean that I have to find moles of O2 and H2?
Answered by DrBob222
You're right about Pgas = X*Ptotal but I don't think you understand the equation I wrote.
Dalton's Law of partial pressures tells us that the total pressure inside a container is the sum of the individual gas pressures. So I wrote the equation
Ptotal = pO2 + pH2O.
That means that the pressure of the oxygen inside the container + the pressure of the H2O inside the container is equal to the total pressure. And the total pressure is given in the problem as 725 torr.
So 725 = pO2 + pH2O
The problem also tells you that the pressure of the H2O at the temperature of the experiment is 25.81 mm (which is the same as 25.81 torr). So you solve this equation for pO2.
725 = pO2 + 25.81
725-25.81 = pO2
pO2 = 699.19 mm Hg or 699.19 Torr. That answer probably will not be counted correct by your prof because it contains too many significant figures. Based on your post of 725 for total P I would round that to 699 mm Hg.
Answered by Natalie
Wow, I feel stupid...I need some sleep. Thanks so much!!
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