$4.4 g m$ of a gas at STP occupies a volume of $2.24$ lit the gas can be:

O_{2}

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C O

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{N O}_{2}

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{C O}_{2}

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Solution

The correct option is A ${C O}_{2}$
At STP,
$2.24 l i t r e$ is ocupied by $4.4 g m$ of gas
Then, $22.4 L$ is occupied by $= \frac{22.4 \times 4.4}{2.24} = 44 g$
Molar mass of $C O_{2}$
$C = 12, O = 16$
$\Rightarrow 12 + 2 (16) \Rightarrow 12 + 32 = 44 g m$
So, the gas will be $44 g$ .

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