Air has three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is mole 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 $40 a m u$ )

1.29

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1.98

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1.6

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2.3

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Solution

The correct option is A 1.29
$1$ mole of sample of air, means total volume= $22.4 l$
$0.78$ mol $N_{2} = 21.84 g m$ $N_{2}$
$0.21$ mol $O_{2} = 6.72 g m$ $O_{2}$
$0.01$ mol $A r = 0.4 g m$ $A r$
Total mass= $28.96 g m$
Density= $\frac{28.96}{22.4} = 1.29 g m / l i t$ .

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Air has three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is mole 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 40amu)

The correct option is A 1.291 mole of sample of air, means total volume= 22.4l0.78 mol N2=21.84gm N20.21 mol O2=6.72gm O20.01 mol Ar=0.4gm ArTotal mass= 28.96gmDensity= 28.9622.4=1.29gm/lit .

Air has three major components: nitrogen, oxygen, and argon. Given that one mole of air at sea level is mole 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 $40 a m u$ )

The correct option is A 1.29
$1$ mole of sample of air, means total volume= $22.4 l$
$0.78$ mol $N_{2} = 21.84 g m$ $N_{2}$
$0.21$ mol $O_{2} = 6.72 g m$ $O_{2}$
$0.01$ mol $A r = 0.4 g m$ $A r$
Total mass= $28.96 g m$
Density= $\frac{28.96}{22.4} = 1.29 g m / l i t$ .