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 s…

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:

Accepted Answer

Correct answer: A. (a) KOH absorbs CO_2. Volume of CO_2 produced =224-64=160 mL Unreacted O_2=64 mL, so O_2 consumed =264-64=200 mL For hydrocarbon C_xH_y : CO_2 Hydrocarbon= 16080=2 ⇒x=2 O_2 consumed per mole =x+ y4= 20080=2.5 2+ y4=2.5 ⇒y=2 Hydrocarbon is C_2H_2

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

Solution

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