The number of molecules of $\mathrm{CO}_2$ liberated by the complete combustion of $0.1 \mathrm{~g}$ atom of…

The number of molecules of $\mathrm{CO}_2$ liberated by the complete combustion of $0.1 \mathrm{~g}$ atom of graphite in air is
  1. $3.01 \times 10^{22}$
  2. $6.02 \times 10^{23}$
  3. $6.02 \times 10^{22}$
  4. $3.01 \times 10^{23}$

Solution

$\underset{1\mathrm{~mol}}{\mathrm{C}(s)}+\mathrm{O}_2(g) \longrightarrow \underset{1\mathrm{~mol}}{\mathrm{CO}_2(g)}$ $=6.023 \times 10^{23}$ $\because 1$ 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 $\mathrm{CO}_2$ $=\frac{6.023 \times 10^{23} \times 0.1}{1}$ $=6.023 \times 10^{22}$

Asked in: AP EAMCET 2010

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