`10` mL of a gaseous organic compound containing `C, H` and `O` only was mixed with `100` mL of `O_(2)` and exploded under condition which allowed the `H_(2)O` formed to condense. The volume of the gas after explosion was `90` mL. On treatment with KOH solution, a further contraction of `20` mL in volume was observed. The vapour density of the compound is `23`. All volume measurements were made under the same condition.
The molecular formula of the compound is

Options

A. `C_(2)H_(4)O`
B. `C_(2)H_(6)O`
C. `C_(3)H_(6)O`
D. None of these

Explanation

To determine the molecular formula of the gaseous organic compound containing carbon (C), hydrogen (H), and oxygen (O), we will follow these steps:

### Step 1: Calculate the volume of O2 that reacted
The initial volume of the gaseous organic compound is 10 mL, and it was mixed with 100 mL of O2. After the explosion, the volume of gas remaining is 90 mL.

**Volume of O2 reacted = Initial volume of O2 - Volume after explosion**
\[ \text{Volume of O2 reacted} = 100 \, \text{mL} - 90 \, \text{mL} = 10 \, \text{mL} \]

### Step 2: Write the balanced chemical reaction
Assuming the organic compound has the formula \( C_xH_yO_z \), the reaction with O2 can be represented as:
\[ C_xH_yO_z + O_2 \rightarrow CO_2 + H_2O \]

### Step 3: Determine the moles of CO2 and H2O produced
From the information given, we know:
- The total volume of gas before the reaction was 10 mL (compound) + 100 mL (O2) = 110 mL.
- After the reaction, the volume is 90 mL, indicating that 20 mL of gas was produced (which is CO2 and H2O).

Since the volume of gases can be directly related to the number of moles (at constant temperature and pressure), we can infer that:
- The volume of CO2 produced = 20 mL (as H2O condenses and does not contribute to the gas volume).

### Step 4: Use the principle of conservation of atoms
For carbon:
\[ x \text{ (from } C_xH_yO_z\text{)} = \text{moles of } CO_2 = 20 \, \text{mL} \]
Thus, \( x = 2 \).

For hydrogen:
Let the volume of H2O produced be \( V_{H2O} \). Since 10 mL of the compound produced 20 mL of CO2, we can relate the moles of H2O to the moles of H in the compound:
\[ \frac{y}{2} = \frac{V_{H2O}}{10} \]

### Step 5: Calculate the volume of H2O produced
From the contraction observed with KOH (20 mL), we can conclude that:
\[ V_{H2O} = 20 \, \text{mL} \]
Thus, using the relation:
\[ y = 2 \times V_{H2O} = 2 \times 20 = 40 \]

### Step 6: Determine the number of oxygen atoms
Using the reaction stoichiometry:
\[ z + \frac{y}{4} = \frac{V_{H2O}}{10} + \frac{V_{O2}}{10} \]
Where \( V_{O2} = 10 \, \text{mL} \) (reacted).
From the equation, we can substitute known values to find \( z \).

### Step 7: Calculate the molecular weight
The vapor density of the compound is given as 23. The molecular weight (M) can be calculated as:
\[ M = 2 \times \text{Vapor Density} = 2 \times 23 = 46 \]

### Step 8: Set up the equation for molecular weight
Using the molecular formula \( C_xH_yO_z \):
\[ 12x + y + 16z = 46 \]
Substituting \( x = 2 \) and \( y = 6 \):
\[ 12(2) + 6 + 16z = 46 \]
\[ 24 + 6 + 16z = 46 \]
\[ 16z = 16 \]
Thus, \( z = 1 \).

### Conclusion
The molecular formula of the compound is:
\[ C_2H_6O \]