Answer:
40.95 L
Explanation:
We'll begin by calculating the number of mole in 12.3 g of O₂. This can be obtained as follow:
Mass of O₂ = 12.3 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mole of O₂ =?
Mole = mass / Molar mass
Mole of O₂ = 12.3 / 32
Mole of O₂ = 0.384 mole
Next, we shall determine the volume occupied by 0.384 mole of O₂. This can be obtained as follow:
Number of mole (n) of O₂ = 0.384 mole
Pressure (P) = 1 atm
Temperature (T) = 273 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) of O₂ =?
PV = nRT
1 × V = 0.384 × 0.0821 × 273
V = 0.384 × 0.0821 × 273
V = 8.6 L
Thus, the volume of O₂ is 8.6 L
Finally, we shall determine the volume of air that contains 8.6 L of O₂. This can be obtained as follow:
Volume of O₂ = 8.6 L
Percentage of O₂ in air = 21%
Volume of air =?
Percentage of O₂ = Vol of O₂ / Vol of air × 100
21% = 8.6 / Vol of air
21 / 100 = 8.6 / Vol of air
Cross multiply
21 × Vol of air = 100 × 8.6
21 × Vol of air = 860
Divide both side by 21
Volume of air = 860 / 21
Volume of air = 40.95 L
Therefore, the volume of air is 40.95 L.
