Question

What volume of “wet” methane would have to be collected at 20.0 oC and 760.0 torr to be sure the sample contained 2.00 x 102 ml of dry methane at the same temperature and pressure ?

My Work: All I have so far is that I use P1V1=P2V2 but I don't know the difference between wet and dry methane?

Answer 1

The dry vs wet means that when you collect a gas over water, the total pressure is what you read on the room manometer and that total pressure is the pressure of the CH4 + the vapor pressure of H2O at that temperature. You can look up the vapor pressure of water at the temperature used.
Ptotal = pCH4 + pH2O. Then plug into that equation and solve for pCH4 which will be the pressure of the dry gas.
So you want 200 mL at 760 mm and you will have x mL at 760-v.p. H2O. Solve for x mL.
Post your work if you get stuck.

Answer 2

So this is what I did
706torr-17.54=742.46

Then plugged into P1V1=P2V2
(742.46)V1=(760)(200)

Solved for V1, and I got 205ml (3 sig figs)
Is this right? Thanks for all your help!