Search any question & find its solution
Question:
Answered & Verified by Expert
An organic compound having molecular formula $\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}$ under goes oxidation with $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}_{2} \mathrm{SO}_{4}$ to produce $\mathrm{X}$ which contains $40 \%$ carbon, $6.7 \%$ hydrogen and $53.3 \%$ oxygen. The molecular formula of the compound $\mathrm{X}$ is
Options:
Solution:
2896 Upvotes
Verified Answer
The correct answer is:
$\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}$
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}-\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{7} / \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}$
$\% \mathrm{C}=\frac{24}{60} \times 100=40 \%, \% \mathrm{H}=\frac{4}{60} \times 100=6.7 \%, \% \mathrm{O}=\frac{32}{60} \times 100=53.3 \%$
$\% \mathrm{C}=\frac{24}{60} \times 100=40 \%, \% \mathrm{H}=\frac{4}{60} \times 100=6.7 \%, \% \mathrm{O}=\frac{32}{60} \times 100=53.3 \%$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.