The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is:
22.414 L mol¯1
It is actually known to several more decimal places but the number above should prove sufficient.This value has been known for about 200 years and it is not a constant of nature like, say, the charge on the electron. If we had picked a different standard temperature, then the molar volume would be different.
Using PV = nRT, you can calculate the value for molar volume. V is the unknown and n = 1.00 mol. Set P and T to their standard values and use R = 0.08206.
(1.00 atm) (V) = (1.00 mol) (0.08206 L atm mol¯1 K¯1) (273 K)V = 22.4 L
Molar volume doesn't show up that often in problems. As a consequence, teachers sometimes like to use molar volume on the test, in order to trip the kid up!! Let's do some examples.
By the way, the problems below can also be solved using PV = nRT. Be aware that using molar volume (like in the problems below) only works at STP. If you have non-standard conditions, you MUST use PV = nRT.
Example #1: you have 2.00 mol of dry H2 at STP. How many liters is this?
Solution:
(2.00 mol) (22,414 L/mol) = 44.8 L (to three sig figs)
Example #2: 0.250 moles of HCl will occupy how many liters at STP?
Solution:
x 22.414 L ––––––– = ––––––– 0.250 mol mol x = (0.250 mol) (22.414 L mol¯1)
x = 5.60 L (to three sig figs)
Example #3: What is molar volume at 576 K?
Solution:
Strictly speaking, this isn't more than a volume-temperature (Charles' Law) problem, but for some reason putting "molar volume" in the problem messes people up.
22.414 L x ––––––– = ––––––– 273 K 546 K x = 44.8 L (to three sig figs)
Notice that mole is not used. This is because the amount of gas (1 mole) never changes.
Example #4: Determine the molar mass of a gas if a 0.250 g sample occupies 0.200 L at STP.
Solution:
0.250 g 22.414 L ––––––– x ––––––– = 28.0 g/mol 0.200 L 1 mole