Some Basic Concepts of Chemistry - Result Question 3
3. At $300 K$ and 1 atmospheric pressure,
$10 mL$ of a hydrocarbon required $55 mL$ of $O _2$ for complete combustion and $40 mL$ of $CO _2$ is formed. The formula of the hydrocarbon is
(2019 Main, 10 April I)
(a) $C _4 H _7 Cl$
(b) $C _4 H _6$
(c) $C _4 H _{10}$
(d) $C _4 H _8$
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Answer:
Correct Answer: 3. (b)
Solution:
In eudiometry,
$\begin{aligned} & \mathrm{C}_x \mathrm{H}_y+\left(x+\frac{y}{4}\right) \mathrm{O}_2 \xrightarrow[1 \mathrm{~atm}]{300 \mathrm{~K}} x \mathrm{CO}_2+\frac{y}{2} \mathrm{H}_2 \mathrm{O} \\ & 1 \mathrm{~mol} \quad\left(x+\frac{y}{4}\right) \mathrm{mol} \quad x \mathrm{~mol} \\ & 1 \mathrm{~mL} \quad\left(x+\frac{y}{4}\right) \mathrm{mL} \quad x \mathrm{~mL} \\ & 10 \mathrm{~mL} \quad\left(x+\frac{y}{4}\right) \times 10 \mathrm{~mL} \quad 10 x \mathrm{~mL} \\ & \end{aligned}$
Given, (i) $V _{CO _2}=10 x=40 mL \Rightarrow x=4$
(ii) $V _{O _2}=10\left(x+\frac{y}{4}\right) mL=55 mL$
$ \begin{aligned} \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{aligned} $
So, the hydrocarbon $\left(C _x H _y\right)$ is $C _4 H _6$.
BETA
