$2.8 imes 10^{-3} \mathrm{~mol}$ of $\mathrm{CO}_{2}$ is left after removing $10^{21}$ molecules f

Question

$2.8 	imes 10^{-3} \mathrm{~mol}$ of $\mathrm{CO}_{2}$ is left after removing $10^{21}$ molecules from its ' $x$ ' mg sample. The mass of $\mathrm{CO}_{2}$ taken initially is Given $N_{A}=6.02 	imes 10^{23} \mathrm{~mol}^{-1}$\

Professor PC Thomas Classes · Answer

(d): Molecules of $\mathrm{CO}_{2}$ left\ $=2.8 	imes 10^{-3} 	imes 6.02 	imes 10^{23}$\ $=1.69 	imes 10^{21}$ molecules\ Initial molecules of $\mathrm{CO}_{2}=$ Number of $\mathrm{CO}_{2}$ molecules left + Number of $\mathrm{CO}_{2}$ molecules removed $=1.69 	imes 10^{21}+10^{21}$ 
$ =2.69 	imes 10^{21} 	ext { molecules } $ 
Initial moles of $\mathrm{CO}_{2}$ 
$ \begin{aligned} & =\frac{2.69 	imes 10^{21}}{6.02 	imes 10^{23}}=4.468 	imes 10^{-3} \mathrm{~mol} \ & 	ext { Mass of } \mathrm{CO}_{2}=	ext { Mole × Molar mass } \ & =4.468 	imes 10^{-3} \mathrm{~mol} 	imes 44 \mathrm{~g} / \mathrm{mol} \ & \left(\because 	ext { Molar mass of } \mathrm{CO}_{2}=44 \mathrm{~g} / \mathrm{mol}ight) \ & =0.196592 \mathrm{~g}=196.6 \mathrm{mg} \end{aligned} $