JEE MAIN - Chemistry (2023 - 8th April Morning Shift - No. 17)
$$0.5 \mathrm{~g}$$ of an organic compound $$(\mathrm{X})$$ with $$60 \%$$ carbon will produce __________ $$\times 10^{-1} \mathrm{~g}$$ of $$\mathrm{CO}_{2}$$ on complete combustion.
Answer
11
Explanation
Mass of Carbon $=12$
Molar Mass of $\mathrm{CO}_2=12+(16 \times 2)=44$
Mass of Compound $=0.5 \mathrm{~g}$
$$ \begin{aligned} &\% \text { of } \mathrm{C} =\frac{\text { Molar mass of } \mathrm{C} \times \text { Mass Of } \mathrm{CO}_2}{\text { Mass Of Compound } \times \text { Molar Mass } \mathrm{Of} ~ \mathrm{CO}_2} \\\\ &\frac{60}{100} =\frac{12 \times x}{0.5 \times 44} \\\\ &1.1 =x \\\\ &x =11 \times 10^{-1} \end{aligned} $$
Molar Mass of $\mathrm{CO}_2=12+(16 \times 2)=44$
Mass of Compound $=0.5 \mathrm{~g}$
$$ \begin{aligned} &\% \text { of } \mathrm{C} =\frac{\text { Molar mass of } \mathrm{C} \times \text { Mass Of } \mathrm{CO}_2}{\text { Mass Of Compound } \times \text { Molar Mass } \mathrm{Of} ~ \mathrm{CO}_2} \\\\ &\frac{60}{100} =\frac{12 \times x}{0.5 \times 44} \\\\ &1.1 =x \\\\ &x =11 \times 10^{-1} \end{aligned} $$
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