JEE MAIN - Chemistry (2023 - 31st January Morning Shift - No. 19)
On complete combustion, $$0.492 \mathrm{~g}$$ of an organic compound gave $$0.792 \mathrm{~g}$$ of $$\mathrm{CO}_{2}$$. The % of carbon in the organic compound is ___________ (Nearest integer)
Answer
44
Explanation
Weight of $\mathrm{C}$ in $0.792 \mathrm{gm} $ $\mathrm{CO}_2$
$$ \begin{aligned} & =\frac{12}{44} \times 0.792=0.216 \\\\ & \% \text { of } C \text { in compound }=\frac{0.216}{0.492} \times 100 \\\\ & =43.90 \% \end{aligned} $$
$$ \begin{aligned} & =\frac{12}{44} \times 0.792=0.216 \\\\ & \% \text { of } C \text { in compound }=\frac{0.216}{0.492} \times 100 \\\\ & =43.90 \% \end{aligned} $$
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