At `200^(@)C PCl_(5)` dissociates as follows :
`PCl_(5)(g0hArrPCl_(3)(g)+Cl_(2)(g)`
It was found that the equilibrium vapours are 62 times as heavy as hydreogen .The degree of dissociation of `PCl_(5)` at `200^(@)C` is nearly :

Options

A. `10%`
B. `42%`
C. `50%`
D. `68%`

Answer

To solve the problem, we need to determine the degree of dissociation of \( PCl_5 \) at \( 200^\circ C \) based on the information provided about the molecular mass of the equilibrium vapors.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

2. **Molecular Mass of Hydrogen**:
The molecular mass of hydrogen (\( H_2 \)) is \( 2 \, g/mol \). Given that the equilibrium vapors are 62 times as heavy as hydrogen, we can calculate the molecular mass of the mixture:
\[ M = 62 \times 2 = 124 \, g/mol \]

3. **Theoretical Molecular Mass of \( PCl_5 \)**:
The theoretical molecular mass of \( PCl_5 \) is known to be \( 208.5 \, g/mol \).

4. **Determining the Number of Moles**:
In the reaction, 1 mole of \( PCl_5 \) dissociates to produce 1 mole of \( PCl_3 \) and 1 mole of \( Cl_2 \). Therefore, if \( \alpha \) is the degree of dissociation, the total moles at equilibrium can be expressed as:
\[ n = 1 - \alpha + (1 - \alpha) + \alpha = 2 - \alpha \]
Here, \( n \) is the total number of moles of gas at equilibrium.

5. **Using the Formula for Average Molecular Mass**:
The average molecular mass of the mixture can be calculated using the formula:
\[ M_{mix} = \frac{M_{PCl_5} - M_{mix}}{n - 1} \times M_{mix} \]
Rearranging gives:
\[ M_{mix} = \frac{M_{PCl_5} - M_{mix}}{(2 - 1)} \Rightarrow M_{mix} = M_{PCl_5} - M_{mix} \]
This leads to:
\[ M_{mix} = \frac{M_{PCl_5}}{1 + \alpha} \]

6. **Substituting Values**:
Now substituting the known values:
\[ 124 = \frac{208.5}{1 + \alpha} \]

7. **Solving for \( \alpha \)**:
Rearranging gives:
\[ 1 + \alpha = \frac{208.5}{124} \]
\[ 1 + \alpha = 1.681 \]
\[ \alpha = 1.681 - 1 = 0.681 \]

8. **Calculating the Degree of Dissociation**:
To express \( \alpha \) as a percentage:
\[ \alpha \times 100 = 0.681 \times 100 = 68.1\% \]

9. **Final Answer**:
The degree of dissociation of \( PCl_5 \) at \( 200^\circ C \) is nearly \( 68\% \).

### Conclusion:
The correct answer is \( D) 68\% \).

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