Chemistry
posted by Clay on .
I'm trying to find the dry gas pressure of H2 in mmHg. The barometric pressure is 765.3 mmHg, the vp of H20 is 23.76 mmHg and the gas collected was 25.6 mL.
this is what i did.
765.3 mmHg  23.76 mmHg  (25.6 Ml/13.6 mmHg) = 739.66 = 740 mmHg 93 sig figs).
Then I have to use the combined gas law to convert this volume of the H2 gas collected to the new volume at STP.
So i used the formula v2 = P1 V1 T2/ T1 P1
I converted the volume of H2 into atm and got .974 atm (740 mmHg/760). Then I multiplied .974 atm by .0244 L (the leftover liquid) and 298 K (the temp.) then I divided by 273 K and 1 atm.
(.974 atm x .0244 L x 298 K)/(273 K x 1 atm) = .0295 L
Is that correct?

The dry gas pressure is pressure inside the tube  watervapor pressure. The pressure has nothing to do with the volume collected, unless you are supporting a column of water, then you have to adjust barometric pressure for the height of the column, it does not appear you had to do this correction.
The rest of it is correct, if dry pressure is corrected.