Volume of 4.4g of $C O_{2}$ at STP is

224 litres

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2.24 litres

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22.4 litres

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44.2 litres

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Solution

The correct option is B
2.24 litres

No. of moles of $C O_{2} = \frac{G i v e n m a s s}{m o l a r m a s s}$

$= \frac{4.4}{44} = 0.1 m o l e s$

No. of moles = $\frac{v o l u m e a v a i l a b l e}{22.4 l i t r e s}$

$\Rightarrow 0.1 = \frac{v o l u m e a v a i l a b l e}{22.4 l i t r e s}$

$\Rightarrow V o l u m e = 0.1 \times 22.4 = 2.24$

Given volume is the required volume here,

So the volume occupied by 4.4g of $C O_{2}$ is 2.24 litres.

Suggest Corrections

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Volume of 4.4g of $C O_{2}$ at STP is

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