The vapour density of the equilibrium mixture of the reaction:
`SO_(2)Cl_(2)(g)hArrSO_(2)(g)+Cl_(2)(g)`
is `50`. The percent dissociation of `SO_(2)Cl_(2)` is

Options

A. `33.00`
B. `35.0`
C. `30.0`
D. `66.00`

Explanation

To solve the problem, we need to find the percent dissociation of \( SO_2Cl_2 \) in the equilibrium mixture given that the vapor density of the mixture is 50.

### Step-by-step Solution:

1. **Understanding Vapor Density**:
The vapor density (VD) is given as 50. We know that the molecular weight (MW) of a gas can be calculated using the formula:
\[ \text{Molecular Weight} = 2 \times \text{Vapor Density} \]
Thus, we can calculate the molecular weight of the equilibrium mixture:
\[ \text{MW} = 2 \times 50 = 100 \]

2. **Molecular Weight of \( SO_2Cl_2 \)**:
The molecular weight of \( SO_2Cl_2 \) can be calculated as follows:
- \( S = 32 \, \text{g/mol} \)
- \( O = 16 \, \text{g/mol} \) (2 atoms)
- \( Cl = 35.5 \, \text{g/mol} \) (2 atoms)
\[ \text{MW of } SO_2Cl_2 = 32 + (2 \times 16) + (2 \times 35.5) = 32 + 32 + 71 = 135 \, \text{g/mol} \]

3. **Using the Dissociation Formula**:
Let \( \alpha \) be the degree of dissociation of \( SO_2Cl_2 \). The reaction is:
\[ SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g) \]
Initially, we have 1 mole of \( SO_2Cl_2 \). At equilibrium, if \( \alpha \) moles dissociate, we will have:
- \( SO_2Cl_2: 1 - \alpha \)
- \( SO_2: \alpha \)
- \( Cl_2: \alpha \)

Thus, the total number of moles at equilibrium is:
\[ n = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

4. **Calculating the Vapor Density of the Equilibrium Mixture**:
The vapor density of the equilibrium mixture can be expressed as:
\[ D = \frac{\text{MW of mixture}}{2} \]
We know that the molecular weight of the mixture can also be expressed in terms of \( \alpha \):
\[ \text{MW of mixture} = \frac{(1 - \alpha) \times 135 + \alpha \times 32 + \alpha \times 35.5}{1 + \alpha} \]
Simplifying this gives:
\[ \text{MW of mixture} = \frac{135 - 135\alpha + 32\alpha + 35.5\alpha}{1 + \alpha} = \frac{135 - 135\alpha + 67.5\alpha}{1 + \alpha} = \frac{135 - 67.5\alpha}{1 + \alpha} \]

5. **Setting Up the Equation**:
We know from our earlier calculation that the molecular weight of the mixture is 100:
\[ \frac{135 - 67.5\alpha}{1 + \alpha} = 100 \]

6. **Solving for \( \alpha \)**:
Cross-multiplying gives:
\[ 135 - 67.5\alpha = 100 + 100\alpha \]
Rearranging terms:
\[ 135 - 100 = 67.5\alpha + 100\alpha \]
\[ 35 = 167.5\alpha \]
\[ \alpha = \frac{35}{167.5} \approx 0.2098 \]

7. **Calculating Percent Dissociation**:
The percent dissociation is given by:
\[ \text{Percent Dissociation} = \alpha \times 100 \approx 0.2098 \times 100 \approx 20.98\% \]

### Final Answer:
The percent dissociation of \( SO_2Cl_2 \) is approximately **21%**.