Density of dry containing only $N_{2}$ and $O_{2}$ is $1.15 g / L$ at $740$ mm of Hg and $300$ K. What is % composition of $N_{2}$ by weight in the air?

78 %

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85.5 %

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70.02 %

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62.75 %

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Solution

The correct option is C $70.02 %$
We know that $P V = n R T$
Since, 1 atm= 760 mm of Hg
$∴$ 740 mm of Hg = 0.97 atm

$M = \frac{d R T}{P} = \frac{1.15 \times 0.082 \times 300}{0.97} = 29.16$ $g$ $m o l^{- 1}$

Let $M_{1}$ be mass fraction of $N_{2}$

$M_{2}$ be that of $O_{2}$ $M_{2} = (1 - M_{1})$

$M_{1} \times 28 + M_{2} \times 32 = 29.16$

On solving $4 W_{1} = 2.83$

$W_{1} = 0.7008$

So percentage of Nitrogen is $70.08$

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density of dry air containing only N2 and O2 is 1.15 g/l at 740mm of hg and 300 k what is the % composition of N2 by mass in the air

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