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 :

Step-by-step solution

KOH absorbs CO₂ . Volume of CO₂ produced = 224 - 64 = 160 mL Unreacted O₂ = 64 mL, so O₂ consumed = 264 - 64 = 200 mL For hydrocarbon C_xH_y : CO₂ Hydrocarbon = 160 80 = 2 x = 2 O₂ consumed per mole = x + y 4 = 200 80 = 2.5 2 + y 4 = 2.5 y = 2 Hydrocarbon is C₂H₂