Air has three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is made up of 78% nitrogen, 21% oxygen, and 1% argon, by volume. What is the density of air? Assume that gases behave in an ideal manner. (Atomic mass of argon is 40 amu)

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Solution

$A v e r a g e a t o m i c m a s s = \frac{x_{1} M_{1} + x_{2} M_{2} + x_{3} M_{3}}{x_{1} + x_{2} + x_{3}} W h e r e x_{1}, x_{2} a n d x_{3} a r e t h e % a b u n d a n c e M_{1,} M_{2} a n d M_{3} a r e t h e m o l a r m a s s A v e r a g e a t o m i c m a s s = \frac{78 \times 28 + 21 \times 32 + 1 \times 40}{78 + 21 + 1} = 28.96 g D e n s i t y = \frac{M a s s}{V o l u m e} = \frac{28.96}{22.4} (\sin c e m o l a r v o l u m e = 22.4 L a t S T P) = 1.293 g / L$

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Air has three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is made up of 78% nitrogen, 21% oxygen, and 1% argon, by volume. What is the density of air? Assume that gases behave in an ideal manner. (Atomic mass of argon is 40 amu)

Average atomic mass = x1M1 + x2M2+ x3M3 x1+x2+ x3Where x1, x2 and x3 are the % abundanceM1,M2 and M3 are the molar massAverage atomic mass = 78 × 28 + 21 × 32+ 1×4078+21+1=28.96 gDensity = MassVolume = 28.9622.4 (since molar volume = 22.4 L at STP) = 1.293 g/L