There are 34 g of oxygen in the container.
We can use the Ideal Gas Law to solve this problem.
[tex]pV = nRT[/tex]
But [tex]n = \frac{m}{M}[/tex], so
[tex]pV = \frac{m}{M}RT[/tex] and
[tex]m = \frac{pVM}{RT}\\[/tex]
STP is 0 °C and 1 bar, so
[tex]m = \frac{\text{1 bar} \times \text{24 L} \times 32.00 \text{ g}\cdot\text{mol}^{-1}}{\text{0.083 14 } \text{bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1}\times\text{273.15 K} } = \textbf{34 g}\\[/tex]
