Composition of Dry AirGas Percent by Volume (%)Nitrogen (N2) 78Oxygen (O2) 21Argon (Ar) 0.9Carbon dioxide (CO2) 0.04 What volume of air contains 12.3 g of oxygen gas at 273 K and 1.00 atm

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.

i believe it is c.) 1575 j.

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