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The number of molecules of $\mathrm{CO}_2$ liberated by the complete combustion of $0.1 \mathrm{~g}$ atom of graphite in air is
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The correct answer is:
$6.02 \times 10^{22}$
$\underset{1\mathrm{~mol}}{\mathrm{C}(s)}+\mathrm{O}_2(g) \longrightarrow \underset{1\mathrm{~mol}}{\mathrm{CO}_2(g)}$
$=6.023 \times 10^{23}$
$\because 1$ mole of graphite on complete combustion gives $\mathrm{CO}_2$
$=6.023 \times 10^{23} \text { molecules }$
$\therefore 0.1$ mole of graphite will give $\mathrm{CO}_2$
$=\frac{6.023 \times 10^{23} \times 0.1}{1}$
$=6.023 \times 10^{22}$
$=6.023 \times 10^{23}$
$\because 1$ mole of graphite on complete combustion gives $\mathrm{CO}_2$
$=6.023 \times 10^{23} \text { molecules }$
$\therefore 0.1$ mole of graphite will give $\mathrm{CO}_2$
$=\frac{6.023 \times 10^{23} \times 0.1}{1}$
$=6.023 \times 10^{22}$
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