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Gay Lussac's Law

0.6 g of carb...

Question

$0.6$ g of carbon was burnt in air to form $C O_{2}$ . Calculate the number of molecules of $C O_{2}$ introduced into the air.

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Solution

0.6 gm of $c = \frac{0.6}{12} = \frac{1}{20} m o l$

$∴$ Molecules of $C O_{2} = \frac{1}{20} \times 6 \times 10^{23}$

$= 3 \times 10^{22}$

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0.6 g

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