How many atoms are contained in 16.80 L of Xe at STP?

1 Answer
Dec 22, 2014

The answer is 4.52 * 10^234.521023 atoms.

This problem can be solved either by using the ideal gas law, PV = nRTPV=nRT, or by using the molar volume of an ideal gas at STP.

If one were to use the ideal gas law, it is important to know that STP means a pressure of 1.00 atm and a temperature of 273.15K. In order to determine the number of atoms, we need to first determine the number of moles

PV = nRT -> n = (PV)/(RT) = (1.00atm * 16.80 L)/(0.082 (L*atm)/(mol*K) * 273.15K) = 0.750 mol esPV=nRTn=PVRT=1.00atm16.80L0.082LatmmolK273.15K=0.750moles

Since 1 mole of any substance has exactly 6.022 * 10^236.0221023 atoms (or molecules - this is known as Avogadro's number), we can determine the number of atoms by

0.750mol es * (6.022 * 10^23at oms)/(1 mol e) = 4.52 * 10^23at oms0.750moles6.0221023atoms1mole=4.521023atoms

The second method uses the fact that 1 mole of any ideal gas occupies 22.4L22.4L at STP. We could determine the number of moles by

n = V/V_(molar) = (16.80L)/(22.4L) = 0.750mol esn=VVmolar=16.80L22.4L=0.750moles

And, once again

0.750mol es * (6.022 * 10^23at oms)/(1 mol e) = 4.52 * 10^23at oms0.750moles6.0221023atoms1mole=4.521023atoms