Question

hydrogen gas is collected over water at a total pressure of 744 mm Hg at 20.0 C. If the volume of the wet hydrogen is 495 mL, what will the dry volume be at standard conditions?

I know that you need to subtract the water vapor pressure from the total and I got 726.45 mmHg. (744mmHg - 17.353mmHg= 726.45 mmHg). I don't know what comes next.

Answer 1

You've done well so far.
Now use (P1V1/T1) = (P2V2/T2) and solve for final volume.
Remember to round the final answer to the correct number of significant figures, expecially if your prof is picky about that kind of thing.

Answer 2

Will the 726.45 mmHg be in place of P1 or P2?

Answer 3

You set it up anyway you wish.
If p1 = 744-17.35 = 726.45 (but I get closer to 726.65)
then v1 = 495 mL
and T1 = 273 + 20 = ?K

If you do it that way then P2 is 760
V2 = ?
T2 = 273
but you could turn all of that around. The point is that it makes little difference if you solve for v2 or v1 as long as you keep the conditions listed together as I've done above.

Answer 4

Thank you so much! This helped me a lot!