Answer:
[tex]\boxed{\text{10.6 L}}[/tex]
Explanation:
Step 1. Calculate the moles of oxygen.
[tex]n = \text{32 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{1.00 mol}[/tex]
Step 2. Calculate the volume of oxygen
We can use the Ideal Gas Law:
pV = nRT
V = nRT/p
(a) Data
p =230 kPa
R = 8.314 kPa·L·K⁻¹mol⁻¹
T = 293 K
(b) Calculation
[tex]V = \dfrac{1.00 \times 8.314 \times 293}{230} = \textbf{10.6 L}\\\\\text{The volume must be }\boxed{\textbf{10.6 L}}[/tex]
