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 will follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of phosphorus pentachloride (PCl5) can be represented as:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

### Step 2: Determine the molecular mass of the equilibrium mixture
We are given that the equilibrium vapors are 62 times as heavy as hydrogen. The molecular mass of hydrogen (H2) is 2 g/mol. Therefore, the molecular mass of the equilibrium mixture is:
\[ \text{Molecular mass of mixture} = 62 \times 2 = 124 \text{ g/mol} \]

### Step 3: Calculate the theoretical molecular mass of the mixture
If no PCl5 dissociates, we only have PCl5 in the gas phase. The molecular mass of PCl5 is:
- P: 31 g/mol
- Cl: 35.5 g/mol (5 Cl atoms)
\[ \text{Molecular mass of PCl}_5 = 31 + (5 \times 35.5) = 31 + 177.5 = 208.5 \text{ g/mol} \]

### Step 4: Set up the equation for degree of dissociation
Let \( \alpha \) be the degree of dissociation of PCl5. At equilibrium, if \( \alpha \) moles of PCl5 dissociate:
- Moles of PCl5 remaining = \( 1 - \alpha \)
- Moles of PCl3 formed = \( \alpha \)
- Moles of Cl2 formed = \( \alpha \)

The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 5: Calculate the average molecular mass of the mixture
The average molecular mass of the mixture can be calculated as:
\[ \text{Average molecular mass} = \frac{\text{Total mass of gas}}{\text{Total moles}} = \frac{(1 - \alpha) \times 208.5 + \alpha \times 120.5}{1 + \alpha} \]
Where 120.5 g/mol is the average mass of PCl3 and Cl2:
\[ \text{Molecular mass of PCl}_3 = 31 + (3 \times 35.5) = 31 + 106.5 = 137.5 \text{ g/mol} \]
\[ \text{Molecular mass of Cl}_2 = 2 \times 35.5 = 71 \text{ g/mol} \]
So, the average molecular mass of the products is:
\[ \text{Average mass} = \frac{137.5 + 71}{2} = 104.25 \text{ g/mol} \]

### Step 6: Set up the equation for degree of dissociation
Now we can set up the equation:
\[ \frac{(1 - \alpha) \times 208.5 + \alpha \times 104.25}{1 + \alpha} = 124 \]

### Step 7: Solve for \( \alpha \)
Cross-multiplying gives:
\[ (1 - \alpha) \times 208.5 + \alpha \times 104.25 = 124 \times (1 + \alpha) \]
Expanding and simplifying:
\[ 208.5 - 208.5\alpha + 104.25\alpha = 124 + 124\alpha \]
Combining like terms:
\[ 208.5 - 124 = 124\alpha + 208.5\alpha - 104.25\alpha \]
\[ 84.5 = (124 + 208.5 - 104.25)\alpha \]
\[ 84.5 = 228.25\alpha \]
\[ \alpha = \frac{84.5}{228.25} \approx 0.370 \]

### Step 8: Convert to percentage
To find the degree of dissociation in percentage:
\[ \text{Degree of dissociation} = \alpha \times 100 \approx 37.0\% \]

### Final Answer
The degree of dissociation of PCl5 at 200°C is nearly **37%**.

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