JEE Main 2016ChemistrySome Basic Concepts of ChemistryMediumMCQ

JEE Main 2016 — Some Basic Concepts of Chemistry Question with Solution

JEE Main 2016 (03 Apr)

Question

At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume, for complete combustion. After combustion, the gases occupy 345 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:
(Assume complete combustion of reactant)

Choose an option

Show full solutionCorrect option: D
Correct answer
DC3H8

Step-by-step explanation

Volume of N2 in air=375×0.8=300 ml
Volume of O2 in air=375×0.2=75 ml
C x H y + x + y 4 O 2 ⟶ xCO 2 g + y 2 H 2 O ℓ 1 5 ml 1 5 x + y 4 0 0 1 5 x -
After combustion, total volume
345=VN2+VCO2
345=300+15x
x=3
Volume of O2 used
15x+y4=75
x+y4=5
y=8
So, hydrocarbon isC3H8.
 

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About this question

This is a previous-year question from JEE Main 2016, covering the Some Basic Concepts of Chemistry chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.