JEE Main 2026 — Jan 21, Shift 1 — Chemistry solution

JEE Main2026Jan 21, Shift 1Chemistry

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 $K O H$ solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:

A
$C_{2} H_{4}$
B
$C_{4} H_{10}$
C
$C_{2} H_{2}$
D
$C_{2} H_{6}$

Solution

$C_{x} H_{y} + (x + y /4) O_{2} \to x C O_{2} + (y /2) H_{2} O$

$O_{2}$ used $= 80 (x + y /4)$ ; $C O_{2}$ formed $= 80 x$ ; residue (excess $O_{2}$ + $C O_{2}$ ) $= 224$ mL.

$K O H$ absorbs $C O_{2} \Rightarrow C O_{2} = 224 - 64 = 160$ mL, so $80 x = 160 \Rightarrow x = 2$ .

$O_{2}$ remaining $= 64 = 264 - 80 (x + y /4) \Rightarrow y = 2$ .

Hydrocarbon $= C_{2} H_{2}$ .

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