Some Basic Concepts of Chemistry 1 Question 3
3. At $300 \mathrm{~K}$ and 1 atmospheric pressure,
$10 \mathrm{~mL}$ of a hydrocarbon required $55 \mathrm{~mL}$ of $\mathrm{O}{2}$ for complete combustion and $40 \mathrm{~mL}$ of $\mathrm{CO}{2}$ is formed. The formula of the hydrocarbon is
(2019 Main, 10 April I)
(a) $\mathrm{C}{4} \mathrm{H}{7} \mathrm{Cl}$
(b) $\mathrm{C}{4} \mathrm{H}{6}$
(c) $\mathrm{C}{4} \mathrm{H}{10}$
(d) $\mathrm{C}{4} \mathrm{H}{8}$
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Solution:
- In eudiometry,
Given, (i) $V_{\mathrm{CO}_{2}}=10 x=40 \mathrm{~mL} \Rightarrow x=4$
(ii) $V_{\mathrm{O}_{2}}=10\left(x+\frac{y}{4}\right) \mathrm{mL}=55 \mathrm{~mL}$
$$ \begin{array}{ll} \Rightarrow & 10\left(4+\frac{y}{4}\right)=55 \ \Rightarrow & 40+\frac{y \times 10}{4}=55 \ \Rightarrow & y \times \frac{10}{4}=15 \Rightarrow y=15 \times \frac{4}{10}=6 \end{array} $$
So, the hydrocarbon $\left(\mathrm{C}{x} \mathrm{H}{y}\right)$ is $\mathrm{C}{4} \mathrm{H}{6}$.
