An organic compound weighing 500 mg , produced 220 mg of $\mathrm{CO}_2$. on complete combustion. The…
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1}$ of $\mathrm{C}: 12, \mathrm{O}: 16$)
Solution
$\begin{aligned} & \mathrm{n}_{\mathrm{CO}_2}=\frac{220 \times 10^{-3}}{44}=5 \times 10^{-3} \mathrm{moles} \\ & \mathrm{m}_{\mathrm{C}}=5 \times 10^{-3} \times 12\end{aligned}$
$\% \mathrm{~m}$ carbon $=\frac{5 \times 10^{-3} \times 12}{500 \times 10^{-3}} \times 100=12 \%$
Correct answer is $12$
Asked in: JEE Main 2025 (07 Apr Shift 1)
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