$2.8 \times 10^{-3} \mathrm{~mol}$ 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: $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
Solution
<p>$$\begin{aligned}
& (\text { moles })_{\text {initial }}=\frac{x \times 10^{-3}}{44} \\
& (\text { moles })_{\text {removal }}=\frac{10^{21}}{6.02 \times 10^{23}} \\
& (\text { moles })_{\text {left }}=(\text { moles })_{\text {initial }}-(\text { moles })_{\text {removed }} \\
& 2.8 \times 10^{-3}=\frac{x \times 10^{-3}}{44}-\frac{10^{21}}{6.02 \times 10^{23}} \\
& \Rightarrow x=196.2 \mathrm{mg}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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