Chemistry · 2026
JEE Main · 21 January 2026, Shift 1 · Q51
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…
$\displaystyle 80$ mL of a hydrocarbon on mixing with $\displaystyle 264$ mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to $\displaystyle 273$ K occupy $\displaystyle 224$ mL . When the system is treated with KOH solution, the volume decreases to $\displaystyle 64$ mL . The formula of the hydrocarbon is :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \mathrm{C}_2 \mathrm{H}_2$
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.