`SO_(3)(g)hArrSO_(2)(g)+(1)/(2)O_(2)(g)`
If observed vapour density of mixture at equilibrium is `35` then find out value of `alpha`

Options

A. `0.28`
B. `0.38`
C. `0.48`
D. `0.58`

Explanation

To solve the problem, we need to find the degree of dissociation (\( \alpha \)) of the reaction:

\[ \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \]

Given that the observed vapor density of the mixture at equilibrium is 35, we can use the formula for degree of dissociation:

\[ \alpha = \frac{D - d}{(n - 1) \cdot d} \]

Where:
- \( D \) = Molar mass of the reactant (SO3) / 2
- \( d \) = Vapor density at equilibrium
- \( n \) = Number of moles of products formed from 1 mole of reactant

### Step 1: Calculate the molar mass of SO3
The molar mass of SO3 can be calculated as follows:

- Molar mass of Sulfur (S) = 32 g/mol
- Molar mass of Oxygen (O) = 16 g/mol

Thus, the molar mass of SO3 is:

\[ \text{Molar mass of SO}_3 = 32 + (16 \times 3) = 32 + 48 = 80 \text{ g/mol} \]

### Step 2: Calculate \( D \)
Now, we can calculate \( D \):

\[ D = \frac{\text{Molar mass of SO}_3}{2} = \frac{80}{2} = 40 \]

### Step 3: Identify \( d \)
The vapor density at equilibrium is given as:

\[ d = 35 \]

### Step 4: Calculate \( n \)
Next, we need to find \( n \). For the reaction, starting with 1 mole of SO3, the products formed are:

- 1 mole of SO2
- 0.5 mole of O2

Thus, the total number of moles of products is:

\[ n = 1 + 0.5 = 1.5 \]

### Step 5: Substitute values into the equation for \( \alpha \)
Now we can substitute the values into the equation for \( \alpha \):

\[ \alpha = \frac{D - d}{(n - 1) \cdot d} = \frac{40 - 35}{(1.5 - 1) \cdot 35} \]

Calculating the numerator:

\[ 40 - 35 = 5 \]

Calculating the denominator:

\[ (1.5 - 1) \cdot 35 = 0.5 \cdot 35 = 17.5 \]

Now substituting these values back into the equation for \( \alpha \):

\[ \alpha = \frac{5}{17.5} = 0.2857 \approx 0.28 \]

### Final Answer
Thus, the value of \( \alpha \) is approximately:

\[ \alpha \approx 0.28 \]