$44\, g\;CO_2 \approx 6 \times 10^{23}$ અણુઓ (લગભગ) 

$\therefore \,\,\,0.2g\,C{O_2}\,\, = \,\,\frac{{6\, \times \,{{10}^{23}}}}{{44}}\, \times \,0.2\,\, = \,\,0.027\, \times \,{10^{23}} \, = \,\,2.72\,\, \times \,{10^{21}}$ અણુ

અહિં $10^{21}$ અણુઓ દુર કરવામાં આવે છે. આથી બાકી અણુઓની સંખ્યા

$= 2.72 \times 10^{21} - 10^{21} = 10^{21} (2.72 - 1) = 1.72 \times 10^{21}$ અણુઓ

અહિં, $6.023 \times 10^{23}$ અણુઓ $= 1$ મોલ 

$\therefore {\text{1}}{\text{.72  }} \times {\text{ 1}}{{\text{0}}^{{\text{21}}}}$ અણુઓ

$ = \,\,\frac{{1\, \times \,1.72\, \times \,{{10}^{21}}}}{{6.023\, \times \,{{10}^{23}}}}\,\, = \,\,2.\,85\, \times \,{10^{ - 3}}$