The volume occupied by 4.4 g of $C O_{2}$ at STP is :

22.4 L

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2.24 L

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0.224 L

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0.1 L

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Solution

The correct option is C 2.24 L
As given,
Mole = $\frac{mass}{Molar mass} = \frac{4.4}{44} = 0.1$

$V_{C O_{2}} = n \times Molar volume = 0.1 \times 22.4 = 2.24$ L

Hence, the correct option is $B$

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