Density of the air

Chemistry Level 2

Air, if the trace gases are neglected, consists essentially of 78% nitrogen, 21% oxygen, and 1% argon (in volume percent).

What is the density ρ \rho of the air when 1 mole of air occupies a volume of V m = 22.4 V_m = 22.4 liters (standard conditions)?

Give the result in units of kg / m 3 \text{kg}/\text{m}^3 and round to the first decimal place.

Note: Air is composed almost entirely of the isotopes 40 Ar , ^{40}\text{Ar}, 16 O , ^{16}\text{O}, and 14 N . ^{14}\text{N}.


The answer is 1.3.

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Mar 18, 2018

The molecules/atoms each have the molar masses M N 2 = 28 g / mol M_{\text{N}_2} = 28 \,\text{g}/\text{mol} , M O 2 = 32 g / mol M_{\text{O}_2} = 32 \,\text{g}/\text{mol} , and M Ar = 40 g / mol M_{\text{Ar}} = 40 \,\text{g}/\text{mol} , so that the density results to ρ = ∑ i c i M i V m = 0.78 ⋅ 28 + 0.21 ⋅ 32 + 0.01 ⋅ 40 22.4 g l ≈ 1.3 kg m 3 \rho = \frac{\sum_{i} c_i M_i}{V_m} = \frac{0.78 \cdot 28 + 0.21 \cdot 32 + 0.01 \cdot 40}{22.4} \,\frac{\text{g}}{\text{l}} \approx 1.3 \, \frac{\text{kg}}{\text{m}^3}

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