`A` mixture of `100ml` of `CO,CO_(2)` and `O_(2)` was sparked. When the resulting gaseous mixture was passed through `KOH` solution, contraction in volume was found to be `80ml`, the composition of initial mixture may be (in the same order)

Options

A. `30ml, 60ml, 10ml`
B. `30ml, 50ml, 20ml`
C. `30ml, 60ml, 20ml`
D. `30ml, 40ml, 30ml`

Explanation

To solve the problem, we need to determine the composition of the initial mixture of gases (CO, CO2, and O2) based on the contraction in volume observed after passing the mixture through KOH solution.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- When the gaseous mixture is sparked, carbon monoxide (CO) reacts with oxygen (O2) to form carbon dioxide (CO2).
- The reaction can be represented as:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
- KOH solution absorbs CO2, leading to a decrease in the volume of the gas mixture.

2. **Volume Contraction**:
- The total volume of the initial gas mixture is 100 ml.
- After the reaction and absorption of CO2 by KOH, the contraction in volume is 80 ml.
- This means that 80 ml of CO2 is absorbed by KOH.

3. **Determining the Amount of CO2**:
- Since KOH absorbs CO2, the volume of CO2 produced from the reaction must equal the contraction observed.
- Therefore, the amount of CO2 produced from the reaction of CO and O2 must be 80 ml.

4. **Setting Up the Equations**:
- Let \( x \) be the volume of CO, \( y \) be the volume of O2, and \( z \) be the volume of CO2 in the initial mixture.
- We know:
\[ x + y + z = 100 \quad \text{(1)} \]
- The volume of CO2 produced from the reaction of CO and O2 can be expressed as:
\[ \text{CO produced} = \frac{x}{2} \quad \text{(from the reaction)} \]
\[ \text{Total CO2 after reaction} = z + \frac{x}{2} = 80 \quad \text{(2)} \]

5. **Solving the Equations**:
- From equation (2):
\[ z + \frac{x}{2} = 80 \quad \Rightarrow \quad z = 80 - \frac{x}{2} \quad \text{(3)} \]
- Substitute equation (3) into equation (1):
\[ x + y + (80 - \frac{x}{2}) = 100 \]
\[ \Rightarrow \quad \frac{x}{2} + y = 20 \quad \text{(4)} \]

6. **Finding Possible Compositions**:
- From equation (4), we can express \( y \) in terms of \( x \):
\[ y = 20 - \frac{x}{2} \]
- We can now test different values of \( x \) to find valid combinations of \( (x, y, z) \) that satisfy all equations and the initial volume constraint.

7. **Testing Options**:
- **Option A**: \( x = 30 \), \( y = 60 \), \( z = 10 \)
- Check:
\[ 30 + 60 + 10 = 100 \quad \text{(valid)} \]
\[ z + \frac{x}{2} = 10 + \frac{30}{2} = 10 + 15 = 25 \quad \text{(not valid)} \]
- **Option B**: \( x = 30 \), \( y = 50 \), \( z = 20 \)
- Check:
\[ 30 + 50 + 20 = 100 \quad \text{(valid)} \]
\[ z + \frac{x}{2} = 20 + \frac{30}{2} = 20 + 15 = 35 \quad \text{(not valid)} \]

- **Option C**: \( x = 30 \), \( y = 60 \), \( z = 20 \)
- Check:
\[ 30 + 60 + 20 = 110 \quad \text{(not valid)} \]

- **Option D**: \( x = 15 \), \( y = 30 \), \( z = 55 \)
- Check:
\[ 15 + 30 + 55 = 100 \quad \text{(valid)} \]
\[ z + \frac{x}{2} = 55 + \frac{15}{2} = 55 + 7.5 = 62.5 \quad \text{(not valid)} \]

### Conclusion:
The valid compositions based on the calculations are \( (30, 50, 20) \) and \( (30, 60, 10) \).