At $0^{\circ} C$ and 1.0 atm $(= 1.01 \times 10^{5} N / m^{2})$ pressure the densities of air, oxygen and nitrogen are $1.284 k g / m^{3}, 1.429 k g / m^{3}$ and $1.251 k g / m^{3}$ respectively. Calculate the percentage of nitrogen in the air from these data, assuming only these two gases to be present.

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Solution

Let us take one mole of air. Let it have $x m o l e$ Nitrogen and $(1 - x) m o l e$ oxygen. Hence,

Since, $ρ = 1.284 = \frac{P M}{R T}$

Where $M = x M_{N} + (1 - x) M_{O}$

Now, $1.429 = \frac{P}{R T} M_{O}$

$1.251 = \frac{P}{R T} M_{N}$

Thus, substituting in above equation,

$1.284 = x (1.251) + (1 - x) (1.429)$

$x = \frac{1.429 - 1.284}{1.429 - 1.251} = \frac{0.145}{0.178} = 0.8146$

Hence, percentage by mass is

$\frac{0.8146 (28)}{0.8146 (28) + 0.1854 (32)} \approx 76.5$

Answer is $76.5$

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At 0∘C and 1.0 atm (=1.01×105N/m2) pressure the densities of air, oxygen and nitrogen are 1.284kg/m3,1.429kg/m3 and 1.251kg/m3 respectively. Calculate the percentage of nitrogen in the air from these data, assuming only these two gases to be present.