JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Some Basic Concepts of Chemistry

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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:

(A) $C_{2} H_{4}$
(B) $C_{4} H_{10}$
(C) $C_{2} H_{2}$ ✓ Correct
(D) $C_{2} H_{6}$

Answer: (C)

Solution

Cooling condenses the water, so the 224 mL of residual gas is CO

_{2}

together with unused O

_{2}

; KOH absorbs the CO

_{2}

and leaves 64 mL, so CO

_{2} = 224 - 64 = 160 mL

and leftover O

_{2} = 64 mL

, giving oxygen consumed

= 264 - 64 = 200 mL

. For C

_{x}

_{y}

, 80 mL of hydrocarbon yields 160 mL of CO

_{2}

, so

x = 160/80 = 2

; and

80 (x + \frac{y}{4}) = 200

gives

2 + \frac{y}{4} = 2.5

, so

y = 2

. The hydrocarbon is C

_{2}

_{2}

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Some Basic Concepts of Chemistry

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