When acethlene (`C_(2)H_(2)`) gas is passed through red hot iron tube, it trimerises into benzene (`C_(6)H_(6)`) vapours .If the average mass of vapours coming out through the tube is 50, calculate the degree of trimerisation of acetylene.

Answer

To solve the problem step by step, we will start by analyzing the reaction and the information provided.

### Step 1: Write the reaction and define variables
The reaction of acetylene (C₂H₂) trimerizing to form benzene (C₆H₆) can be represented as:
\[ 3 \, \text{C}_2\text{H}_2 \rightarrow \text{C}_6\text{H}_6 \]
Let \( \alpha \) be the degree of trimerization of acetylene. Initially, we have 1 mole of acetylene and 0 moles of benzene.

### Step 2: Determine moles at equilibrium
At equilibrium, the moles of acetylene and benzene can be expressed as:
- Moles of C₂H₂ = \( 1 - \alpha \)
- Moles of C₆H₆ = \( \frac{\alpha}{3} \)

### Step 3: Calculate the average molar mass of the mixture
The average mass of the vapors is given as 50 g. The average molar mass (\( M_{avg} \)) of the mixture can be calculated using the formula:
\[ M_{avg} = \frac{(\text{moles of C}_2\text{H}_2 \times M_{C_2H_2}) + (\text{moles of C}_6\text{H}_6 \times M_{C_6H_6})}{\text{total moles}} \]
Where:
- \( M_{C_2H_2} = 26 \, \text{g/mol} \)
- \( M_{C_6H_6} = 78 \, \text{g/mol} \)

### Step 4: Substitute values into the average mass equation
Substituting the values into the equation:
\[ 50 = \frac{(1 - \alpha) \cdot 26 + \left(\frac{\alpha}{3}\right) \cdot 78}{(1 - \alpha) + \frac{\alpha}{3}} \]

### Step 5: Simplify the equation
The total moles in the denominator can be simplified:
\[ (1 - \alpha) + \frac{\alpha}{3} = 1 - \alpha + \frac{\alpha}{3} = 1 - \frac{3\alpha}{3} + \frac{\alpha}{3} = 1 - \frac{2\alpha}{3} \]
Now substituting this back into the equation:
\[ 50 = \frac{(1 - \alpha) \cdot 26 + \left(\frac{\alpha}{3}\right) \cdot 78}{1 - \frac{2\alpha}{3}} \]

### Step 6: Cross-multiply and solve for \( \alpha \)
Cross-multiplying gives:
\[ 50 \left(1 - \frac{2\alpha}{3}\right) = (1 - \alpha) \cdot 26 + \left(\frac{\alpha}{3}\right) \cdot 78 \]
Expanding both sides:
\[ 50 - \frac{100\alpha}{3} = 26 - 26\alpha + 26\alpha \]
This simplifies to:
\[ 50 - \frac{100\alpha}{3} = 26 + 26\alpha \]
Rearranging gives:
\[ 50 - 26 = \frac{100\alpha}{3} + 26\alpha \]
\[ 24 = \frac{100\alpha}{3} + \frac{78\alpha}{3} \]
\[ 24 = \frac{178\alpha}{3} \]
Multiplying through by 3:
\[ 72 = 178\alpha \]
Thus,
\[ \alpha = \frac{72}{178} \approx 0.4045 \]

### Step 7: Conclusion
The degree of trimerization \( \alpha \) is approximately 0.4045 or 40.45%.