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# Volume occupied by 4.4g of $CO_2$ is

$\begin{array}{1 1}(a)\;224litres& (b)\;2.24litres \\(c)\;22.4\times 10^2litres&(d)\;22.4litres\end{array}$

Molecular weight of $CO_2$ =44
44g of $CO_2$ occupy $22.4$litres
4.4g of $CO_2$ will occupy $\large\frac{22.4}{44}$$\times 4.4$
$\Rightarrow 2.24$litres
Hence (b) is the correct answer.