$2.8\times {10}^{-3}$ mol of CO₂ is left after removing 10²¹ molecules from its ' x ' mg sample. The mass of CO₂ taken initially is
Given : ${\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{\mathrm{mol}}^{-1}$

$\begin{array}{l}(\mathrm{moles}{)}_{\text{initial }}=\frac{\mathrm{x}\times {10}^{-3}}{44}\\ (\mathrm{moles}{)}_{\text{removal }}=\frac{{10}^{21}}{6.02\times {10}^{23}}\\ (\text{ moles }{)}_{\text{left }}=(\text{ moles }{)}_{\text{initial }}-(\text{ moles }{)}_{\text{removed }}\\ 2.8\times {10}^{-3}=\frac{\mathrm{x}\times {10}^{-3}}{44}-\frac{{10}^{21}}{6.02\times {10}^{23}}\\ \Rightarrow x=196.2\mathrm{mg}\end{array}$