80 mL of an organic compound combines with 264 mL O₂ and on ignition gives 224 mL of gaseous mixture at NTP. After passing through KOH, 64 mL gas remains.

The organic compound is:

The correct option is C₂H₂.

Step-by-step Derivation and Explanation:

Let the molecular formula of the organic compound be represented as $C_x{H}_y$.

The general balanced equation for the combustion of a hydrocarbon $C_x{H}_y$ in oxygen is:
$C_x{H}_y+(x+\frac{y}{4})O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$

According to Gay-Lussac's Law of Gaseous Volumes, the volumes of gases react in simple whole-number ratios under the same conditions of temperature and pressure. Therefore, we can relate the volumes directly to the stoichiometric coefficients:
1 volume of $C_x{H}_y$ reacts with $(x+\frac{y}{4})$ volumes of $O_2$ to produce $x$ volumes of $C{O}_2$ gas (since water is in the liquid state at NTP and its volume is negligible compared to the gas volumes).

Given parameters:
- Volume of organic compound, $V_{\mathrm{hydrocarbon}}=80\text{mL}$
- Total volume of $O_2$ supplied = 264 mL

After ignition, the gaseous mixture (which contains produced $C{O}_2$ and any unreacted, excess $O_2$) has a volume of 224 mL.
When this gaseous mixture is passed through a KOH solution, $C{O}_2$ is absorbed completely:
$2\mathrm{KOH}+C{O}_2\to K_{2}C{O}_3+H_{2}O$

The remaining gas is unreacted (excess) $O_2$, which is given as 64 mL.
Therefore:
- Volume of unreacted $O_2=64\text{mL}$
- Volume of $C{O}_2$ produced = Total volume of gaseous mixture - Volume of remaining gas
$V_{C{O}_2}=224\text{mL}-64\text{mL}=160\text{mL}$

Now, let's find the volume of $O_2$ consumed during combustion:
- Volume of $O_2$ consumed = Total $O_2$ supplied - Unreacted $O_2$
$V_{O_2\text{consumed}}=264\text{mL}-64\text{mL}=200\text{mL}$

Using stoichiometry:
1. Finding $x$ (number of carbon atoms):
From the reaction, 80 mL of $C_x{H}_y$ produces $80x\text{mL}$ of $C{O}_2$.
$80x=160$
$x=\frac{160}{80}=2$

2. Finding $y$ (number of hydrogen atoms):
From the reaction, 80 mL of $C_x{H}_y$ consumes $80(x+\frac{y}{4})\text{mL}$ of $O_2$.
$80(x+\frac{y}{4})=200$
Substitute $x=2$ into the equation:
$80(2+\frac{y}{4})=200$
$2+\frac{y}{4}=\frac{200}{80}$
$2+\frac{y}{4}=2.5$
$\frac{y}{4}=0.5$
$y=0.5\times 4=2$

Substituting the values of $x$ and $y$ back into the molecular formula $C_x{H}_y$, we get C₂H₂ (Acetylene).