A closed container of volume $30 l i t r e$ contains a mixture of nitrogen and oxygen gases, at a tempereture of $27$ and pressure of $4 a t m .$ The total mass of the mixture is $148 g m .$ The moles of individual gases in the container are (Take $R = 0.08 l i t r e a t m / m o l e K$ )

n_{N_{2}} = 2 m o l e s, n_{O_{2}} = 3 m o l e

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n_{N_{2}} = 3 m o l e, n_{O_{2}} = 2 m o l e

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n_{N_{2}} = 4 m o l e, n_{O_{2}} = 1 m o l e

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n_{N_{2}} = 2.5 m o l e, n_{O_{2}} = 2.5 m o l e

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Solution

The correct option is D $n_{N_{2}} = 3 m o l e, n_{O_{2}} = 2 m o l e$
$p = 4 a t m$ , $v = 30 l i t r e s$ , $T = 300 K$ , $R = 0.08 l i t r e a t m / m o l e K$
By ideal gas equation,
$p v = n_{T} R T$
$∴ n_{T} = \frac{p v}{R T} = \frac{4 \times 30}{0.08 \times 300} = 5$
$∴ n_{N_{2}} + n_{O_{2}} = 5$
Let, $x$ be the weight of $N_{2}$ in grams
$∴$ weight of $O_{2} = (148 - x)$ grams
$∴ \frac{x}{28} + \frac{(148 - x)}{32} = 5$
$∴ 32 x + 4144 - 28 x = 4480$
$∴ 4 x = 336$
$∴ x = 84 g$
$∴$ No. of moles of $N_{2} (n_{{N O}_{2}}) = \frac{84}{28} = 3$ moles
& No. of moles of $O_{2} (n_{O_{2}}) = \frac{148 - 84}{32} = 2$ moles

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