For the equlibrium `SO_(3)(g)hArrSO_(2)(g)+1/2O(2)(g)` the molar mass at equlibrium was observed to be 60. then the degree of dissociation of `SO_(3)` would be

Options

A. `0.33`
B. `0.66`
C. `0.25`
D. `0.50`

Answer

To find the degree of dissociation (α) of \( SO_3 \) in the equilibrium reaction

\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \]

given that the molar mass at equilibrium is 60 g/mol, we can follow these steps:

### Step 1: Write the Initial and Equilibrium Conditions
Initially, we have 1 mole of \( SO_3 \). At equilibrium, let the degree of dissociation of \( SO_3 \) be \( \alpha \).

- Initial moles:
- \( SO_3 \): 1
- \( SO_2 \): 0
- \( O_2 \): 0

- At equilibrium:
- \( SO_3 \): \( 1 - \alpha \)
- \( SO_2 \): \( \alpha \)
- \( O_2 \): \( \frac{\alpha}{2} \)

### Step 2: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium can be calculated as follows:

\[ \text{Total moles} = (1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} \]

### Step 3: Calculate the Molar Mass at Equilibrium
The molar mass at equilibrium can be calculated using the formula for average molar mass based on the total moles:

\[ \text{Average molar mass} = \frac{\text{Total mass}}{\text{Total moles}} \]

The total mass at equilibrium is given by:

\[ \text{Total mass} = (1 - \alpha) \cdot M_{SO_3} + \alpha \cdot M_{SO_2} + \frac{\alpha}{2} \cdot M_{O_2} \]

Where:
- \( M_{SO_3} = 80 \, \text{g/mol} \)
- \( M_{SO_2} = 64 \, \text{g/mol} \)
- \( M_{O_2} = 32 \, \text{g/mol} \)

Thus, the total mass becomes:

\[ \text{Total mass} = (1 - \alpha) \cdot 80 + \alpha \cdot 64 + \frac{\alpha}{2} \cdot 32 \]

This simplifies to:

\[ \text{Total mass} = 80 - 80\alpha + 64\alpha + 16\alpha = 80 - 80\alpha + 80\alpha = 80 \]

### Step 4: Set Up the Equation for Molar Mass
At equilibrium, we know the molar mass is 60 g/mol. Therefore, we can set up the equation:

\[ \frac{80}{1 + \frac{\alpha}{2}} = 60 \]

### Step 5: Solve for α
Cross-multiplying gives:

\[ 80 = 60 \left(1 + \frac{\alpha}{2}\right) \]

Expanding this gives:

\[ 80 = 60 + 30\alpha \]

Rearranging gives:

\[ 20 = 30\alpha \]

Thus,

\[ \alpha = \frac{20}{30} = \frac{2}{3} \approx 0.67 \]

### Conclusion
The degree of dissociation \( \alpha \) of \( SO_3 \) is approximately \( 0.67 \) or \( \frac{2}{3} \).