JEE Chemistry 2026 Question with Solution

Given: Volume of hydrocarbon = $80 \, \text{mL}$, initial oxygen = $264 \, \text{mL}$, volume after combustion and cooling = $224 \, \text{mL}$, volume after treatment with KOH = $64 \, \text{mL}$.

Find: The formula of the hydrocarbon.

KOH absorbs $CO_2$, so the gas left after KOH treatment is residual $O_2$.

Therefore, volume of $CO_2$ is

Volume of $O_2$ reacted is

Let the hydrocarbon be $C_xH_y$.

The combustion reaction is

By Gay-Lussac's law, gas volumes are proportional to stoichiometric coefficients.

From $CO_2$ formed:

Now use oxygen consumed:

Hence, the hydrocarbon is $C_2H_2$.

Therefore, the correct option is C.

Given: The hydrocarbon volume is $80 \, \text{mL}$.

Find: Values of $x$ and $y$ in $C_xH_y$.

Since KOH removes $CO_2$, the decrease in volume due to KOH gives the volume of $CO_2$ absorbed:

So $80 \, \text{mL}$ of hydrocarbon produces $160 \, \text{mL}$ of $CO_2$. Hence

So one volume of hydrocarbon produces $2$ volumes of $CO_2$, giving

Also, oxygen consumed is

Thus $80 \, \text{mL}$ of hydrocarbon consumes $200 \, \text{mL}$ of oxygen, so

Hence one volume of hydrocarbon consumes $2.5$ volumes of $O_2$.

For $C_xH_y$,

Substituting $x = 2$,

Therefore, the formula is $C_2H_2$, so the correct option is C.