What would be the volume of a 13.70 g sample of the noble gas krypton at STP?

1 Answer
Mar 28, 2015

The volume occupied by the gas will be #"3.711 L"#.

When you are given STP conditions, which imply a pressure of 100 kPa and a temperature of -273.15 K, you must think molar volume of a gas.

At STP, 1 mole of any ideal gas occupies exactly 22.7 L. This means that, in order to get the volume of a gas at STP, all you need to do is figure out how many moles of the gas you're dealing with.

Use krypton's molar mass to determine how many moles you have in 13.70 g

#"13.70"cancel("g") * "1 mole Kr"/("83.798"cancel("g")) = "0.1635 moles Kr"#

So, if 1 mole occupies 22.7 L, fewer moles will of course occupy a smaller volume.

#"0.1635"cancel("moles") * "22.7 L"/cancel("1 mole") = color(red)("3.711 L")#

SIDE NOTE You'll very often see the molar volume of a gas at STP being given as 22.4 L - that's the (very) old value used for STP conditions defined as 1 atm and 0 degrees Celsius.

I suggest using 22.7 L, the actual value, unless your teacher instructs you to use 22.4 L.