$2.8 \times 10^{-3} \mathrm{~mol}$ of $\mathrm{CO}_2$ is left after removing $10^{21}$ molecules from its '…

$2.8 \times 10^{-3} \mathrm{~mol}$ 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: $\mathrm{N}_{\mathrm{A}}=6.02 \times 10^{23} \mathrm{~mol}^{-1}$
  1. 98.3 mg
  2. 48.2 mg
  3. 196.2 mg
  4. 150.4 mg

Solution

Moles of removed $\mathrm{CO}_2=\frac{10^{21}}{6.02 \times 10^{23}} \mathrm{~mol}$
$=1.66 \times 10^{-3} \mathrm{~mol}$
mole of $\mathrm{CO}_2$ left $=2.8 \times 10^{-3} \mathrm{moles}$ total moles of $\mathrm{CO}_2$ taken initially
$=(2.8+1.66) \times 10^{-3} \mathrm{~mol}$
mass of $\mathrm{CO}_2$ taken initially
$\begin{aligned}
& =4.46 \times 10^{-3} \times 44 \\
& =196.24 \times 10^{-3} \mathrm{~g} \\
& =196.24 \mathrm{mg}
\end{aligned}$

Asked in: JEE Main 2025 (23 Jan Shift 1)

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