A mixture of `C_(2)H_(2)` and `C_(3)H_(8)` occupied a certain volume at `80mm` Hg. The mixture was completely burnt to `CO_(2)` and `H_(2)O` (I). When the pressure of `CO_(2)` was found to be `230mm` Hg at the same temperature and volume, the fraction of `C_(2)H_(2)` in mixture is .

Options

A. `0.125`
B. `0.5`
C. `0.85`
D. `0.25`

Explanation

To solve the problem step by step, we need to determine the fraction of `C₂H₂` (ethyne) in the mixture of `C₂H₂` and `C₃H₈` (propane) based on the given pressures and combustion reactions.

### Step 1: Define Variables
Let the partial pressure of `C₂H₂` be \( x \) mm Hg. Therefore, the partial pressure of `C₃H₈` will be \( 80 - x \) mm Hg, since the total pressure of the mixture is 80 mm Hg.

### Step 2: Write the Balanced Combustion Reaction
The balanced combustion reaction for `C₂H₂` and `C₃H₈` can be written as:
- For `C₂H₂`:
\[ C₂H₂ + 5O₂ \rightarrow 2CO₂ + 2H₂O \]
- For `C₃H₈`:
\[ C₃H₈ + 5O₂ \rightarrow 3CO₂ + 4H₂O \]

### Step 3: Calculate the Pressure of `CO₂`
From the combustion reactions, we can determine how much `CO₂` is produced from each component:
- From `C₂H₂`, the pressure of `CO₂` produced is \( 2x \) (since 1 mole of `C₂H₂` produces 2 moles of `CO₂`).
- From `C₃H₈`, the pressure of `CO₂` produced is \( 3(80 - x) \) (since 1 mole of `C₃H₈` produces 3 moles of `CO₂`).

Thus, the total pressure of `CO₂` produced is:
\[ P_{CO₂} = 2x + 3(80 - x) \]

### Step 4: Set Up the Equation
According to the problem, the pressure of `CO₂` after combustion is 230 mm Hg. Therefore, we can set up the equation:
\[ 2x + 3(80 - x) = 230 \]

### Step 5: Solve for \( x \)
Now, simplify and solve the equation:
\[ 2x + 240 - 3x = 230 \]
\[ 240 - x = 230 \]
\[ -x = 230 - 240 \]
\[ -x = -10 \]
\[ x = 10 \]

### Step 6: Calculate the Fraction of `C₂H₂`
Now that we have \( x = 10 \) mm Hg (the pressure of `C₂H₂`), we can find the fraction of `C₂H₂` in the mixture:
\[ \text{Fraction of } C₂H₂ = \frac{x}{80} = \frac{10}{80} = 0.125 \]

### Conclusion
The fraction of `C₂H₂` in the mixture is \( 0.125 \).