The number of molecules of ${CO}_2$ liberated by the complete combustion of 0.1 gram atom of graphite in air is

Given, Weight of ${CO}_2$ is 0.1 gram.

$$\begin{gathered} \mathrm{C} + {\mathrm{O}}_2 \to {\mathrm{CO}}_2 \\ 1\mathrm{mole} 1\mathrm{mole} \\ 1\mathrm{gm}-\mathrm{atom} 1\mathrm{gm}-\mathrm{atom} \end{gathered}$$

 1 gram atom contains = $6.02\times {10}^{23}$ molecules

So, 0.1 gram of atom contains $6.02\times {10}^{22}$ molecules.

Therefore, option C is correct.