A mixture of a mol of `C_(3) H_(8)` and `b` mol of `C_(2) H_(4)` was kept is a container of `V L` exerts a pressure of 4.93 atm at temperature `T`. Mixture was burnt in presence of `O_(2)` to convert `C_(3) H_(8)` and `C_(2) H_(4)` into `CO_(2)` in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilirium with atomsphere at temperature `T` was found to be 11.08 atm.
The moles fraction of `C_(3) H_(8)` in the mixture is

Options

A. 0.25
B. 0.75
C. 0.45
D. 0.55

Explanation

To solve the problem step by step, we need to analyze the information given and apply the ideal gas law and stoichiometry of the combustion reactions.

### Step 1: Understand the Initial Conditions
We have a mixture of:
- 1 mole of propane (C₃H₈)
- b moles of ethylene (C₂H₄)

The total pressure of this mixture in a container of volume V at temperature T is given as 4.93 atm.

Using the ideal gas law:
\[ P = \frac{nRT}{V} \]
where:
- \( P \) = pressure
- \( n \) = total number of moles
- \( R \) = universal gas constant
- \( T \) = temperature
- \( V \) = volume

### Step 2: Write the Initial Pressure Equation
The total number of moles \( n \) in the mixture is \( 1 + b \). Therefore, the equation for the initial pressure can be written as:
\[ 4.93 = \frac{(1 + b)RT}{V} \]
This is our **Equation 1**.

### Step 3: Write the Combustion Reactions
The combustion reactions for propane and ethylene are as follows:
1. For propane (C₃H₈):
\[ C₃H₈ + 5O₂ \rightarrow 3CO₂ + 4H₂O \]
This means 1 mole of C₃H₈ produces 3 moles of CO₂.

2. For ethylene (C₂H₄):
\[ C₂H₄ + 3O₂ \rightarrow 2CO₂ + 2H₂O \]
This means 1 mole of C₂H₄ produces 2 moles of CO₂.

### Step 4: Calculate Moles of CO₂ Produced
After combustion, the total moles of CO₂ produced from the reaction will be:
- From 1 mole of C₃H₈: 3 moles of CO₂
- From b moles of C₂H₄: \( 2b \) moles of CO₂

Thus, the total moles of CO₂ after combustion:
\[ n_{CO₂} = 3 \times 1 + 2 \times b = 3 + 2b \]

### Step 5: Write the Final Pressure Equation
After combustion, the pressure of the gases in the container is given as 11.08 atm. The total moles of gas now is \( 3 + 2b \) (since water vapor is not counted in the pressure). Therefore, we can write:
\[ 11.08 = \frac{(3 + 2b)RT}{V} \]
This is our **Equation 2**.

### Step 6: Set Up the Ratio of Equations
Now, we can set up a ratio from Equation 1 and Equation 2:
\[ \frac{11.08}{4.93} = \frac{3 + 2b}{1 + b} \]

### Step 7: Solve for b
Cross-multiplying gives:
\[ 11.08(1 + b) = 4.93(3 + 2b) \]
Expanding both sides:
\[ 11.08 + 11.08b = 14.79 + 9.86b \]
Rearranging gives:
\[ 11.08b - 9.86b = 14.79 - 11.08 \]
\[ 1.22b = 3.71 \]
\[ b = \frac{3.71}{1.22} \approx 3.04 \]

### Step 8: Calculate Mole Fraction of C₃H₈
Now that we have \( b \), we can find the mole fraction of C₃H₈:
\[ \text{Mole fraction of } C₃H₈ = \frac{1}{1 + b} = \frac{1}{1 + 3.04} = \frac{1}{4.04} \approx 0.247 \]

### Final Answer
The mole fraction of C₃H₈ in the mixture is approximately **0.25**.