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 so

Question

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) C2H4
(B) C4H10
(C) C2H2
(D) C2H6

Prashant Kotian · Accepted Answer

Correct answer: C. x y(g) 2(g) 2(g) ( ) y y C H x O xCO H O 2           t=0 80 264 0 - t=tfinal - y 80 x         80x - y 80 x 80x 4           80y 224   80y y 4    y 80 x 4          264 80 x 2          264 – 80x – 40 = 64 x = 2 ALLEN 3 JEE-Main Exam Session-1 (January 2026)/21- 

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