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

Dec 22, 2014

The answer is $4.52 \cdot {10}^{23}$ atoms.

This problem can be solved either by using the ideal gas law, $P V = n R T$, 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

$P V = n R T \to n = \frac{P V}{R T} = \frac{1.00 a t m \cdot 16.80 L}{0.082 \frac{L \cdot a t m}{m o l \cdot K} \cdot 273.15 K} = 0.750 m o l e s$

Since 1 mole of any substance has exactly $6.022 \cdot {10}^{23}$ atoms (or molecules - this is known as Avogadro's number), we can determine the number of atoms by

$0.750 m o l e s \cdot \frac{6.022 \cdot {10}^{23} a t o m s}{1 m o l e} = 4.52 \cdot {10}^{23} a t o m s$

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

$n = \frac{V}{V} _ \left(m o l a r\right) = \frac{16.80 L}{22.4 L} = 0.750 m o l e s$

And, once again

$0.750 m o l e s \cdot \frac{6.022 \cdot {10}^{23} a t o m s}{1 m o l e} = 4.52 \cdot {10}^{23} a t o m s$