Q9 / 160 · TS EAMCET 2022 (19 Jul Shift 2) · Chemistry
Combustion of $10 \mathrm{ml}$ of a gaseous hydrocarbon gives $40 \mathrm{ml}$ of $\mathrm{CO}_2$ and $50 \mathrm{ml}$ of water vapour under the same conditions. The molecular formula of the hydrocarbon is
- A$\mathrm{C}_4 \mathrm{H}_6$
- B$\mathrm{C}_4 \mathrm{H}_8$
- C$\mathrm{C}_3 \mathrm{H}_4$
- Correct answerD$\mathrm{C}_4 \mathrm{H}_{10}$
Solution
Ratio of volume $=10: 40: 50=1: 4: 5$
In hydrocarbon, all carbon atoms are from $\mathrm{CO}_2 10 \mathrm{~mL}$ of hydrocarbon produce $40 \mathrm{~mL}$ of $\mathrm{CO}_2=1: 4$ $1 \mathrm{~mol}$ of hydrocarbon has $4 \mathrm{~mol}$ of $\mathrm{C}$. $10 \mathrm{~mL}$ of hydrocarbon produce $50 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}=1: 5$ Each water molecule has $2 \mathrm{~mol}$ of $\mathrm{H}$ atom. So, $1 \mathrm{~mol}$ hydrogen carbon compound has 10 moles of hydrogen which gives $\mathrm{C}_4 \mathrm{H}_{10}$.
Hence, the correct answer is option: 'D'.