Some Basic Concepts of Chemistry - Result Question 13
14. At 300 K and $1 \mathrm{~atm}, 15 \mathrm{~mL}$ of a gaseous hydrocarbon requires 375 mL air containing $20 % \mathrm{O}_2$ by volume for complete combustion. After combustion, the gases occupy 330 mL . Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is
(2016 JEE Main)
(a) $\mathrm{C}_3 \mathrm{H}_8$
(b) $\mathrm{C}_4 \mathrm{H}_8$
(c) $\mathrm{C}4 \mathrm{H}{10}$
(d) $\mathrm{C}_3 \mathrm{H}_6$
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Answer:
Correct Answer: 14. (*)
Solution:
$\begin{aligned} & \mathrm{O}_2 \text { used } \quad=20 % \text { of } 375=75 \mathrm{~mL} \\ & \text { Inert part of air }=80 % \text { of } 375=300 \mathrm{~mL} \\ & \text { Total volume of gases }=\mathrm{CO}_2+\text { Inert part of air } \\ & =30+300=330 \mathrm{~mL} \\ & \frac{x}{1}=\frac{30}{15} \Rightarrow x=2 \\ & \frac{x+\frac{y}{4}}{1}=\frac{75}{15} \Rightarrow x+\frac{y}{4}=5 \\ & \Rightarrow \quad x=2, y=12 \Rightarrow \mathrm{C}2 \mathrm{H}{12} \\ & \end{aligned}$
BETA
