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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2026 · 21 January · Morning Shift · Chemistry · Some Basic Concepts of Chemistry

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed...

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

Options

  1. AC2H2\mathrm{C}_{2} \mathrm{H}_{2}
  2. BC2H4\mathrm{C}_{2} \mathrm{H}_{4}
  3. CC2H6\mathrm{C}_{2} \mathrm{H}_{6}
  4. DC4H10\mathrm{C}_{4} \mathrm{H}_{10}

Correct answer

A. C2H2\mathrm{C}_{2} \mathrm{H}_{2}

Step-by-step solution

KOH absorbs CO2CO_2. Volume of CO2CO_2 produced = 224−64=160224 - 64 = 160 mL
Unreacted O2=64O_2 = 64 mL, so O2O_2 consumed = 264−64=200264 - 64 = 200 mL
For hydrocarbon CxHyC_xH_y: CO2Hydrocarbon=16080=2⇒x=2\frac{CO_2}{\text{Hydrocarbon}} = \frac{160}{80} = 2 \Rightarrow x = 2
O2O_2 consumed per mole = x+y4=20080=2.5x + \frac{y}{4} = \frac{200}{80} = 2.5
2+y4=2.5⇒y=22 + \frac{y}{4} = 2.5 \Rightarrow y = 2
Hydrocarbon is C2H2C_2H_2
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