Explanation

(c)
Molecular mass of $CO_{2} = 12 + 32 = 44$
$44g$ of $CO_{2}$ has = 6.023$\times 10^{23}$ molecule
$0.2g$ of $CO_{2}$ has $= \frac{6.023 \times 10^{23}}{44} \times 0.2$
$= 0.0273 \times 10^{23}$
If $10^{21}$ molecules are removed then number of molecules
$= 1.73 \times 10^{21}$
$\because 6.023 \times 10^{23}$ molecules = 1 mol
$\therefore 1.73 \times 10^{21}$ molecules $= \frac{1}{6.023 \times 10^{23}} \times 1.73 \times 10^{21}$
$= 0.0028$ mol