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Question #1698b

1 Answer

Aditya Banerjee.

Nov 10, 2016

$4.48 L$ $4.48 L$ of $C O_{2}$ $CO_2$ contains $1.2064 \times 10^{23}$ $1.2064xx10^(23)$ molecules.

Explanation:

By, Avogadro's Law,
$22.40 L$ $22.40 L$ of $C O_{2}$ $CO_2$ contains $6.023 \times 10^{23}$ $6.023xx10^(23)$ molecules.

$:.4.48 L$ of $CO_2$ contains $((6.023xx10^(23))xx4.48)/22.4$ molecules.

$:.4.48 L$ of $CO_2$ contains $1.2064xx10^(23)$ molecules. (answer).

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