Phosphorous pentachloride when heated in a sealed tube at `700 K` it undergoes decomposition as
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g), K_(p)=38` atm
Vapour density of the mixture is `74.25`.
Percentage dissociation of `PCl_(5)` may be given as

Options

A. `4.04`
B. `40.4`
C. `44.0`
D. `0.404`

Answer

To solve the problem of finding the percentage dissociation of phosphorus pentachloride (PCl₅) when heated, we can follow these steps:

### Step-by-Step Solution:

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

2. **Given Data**:
- \( K_p = 38 \, \text{atm} \)
- Vapour density of the mixture = 74.25

3. **Calculate the Molecular Weight of the Mixture**:
The vapour density (VD) is related to the molecular weight (M) by the formula:
\[ \text{VD} = \frac{M}{2} \]
Therefore, we can calculate the molecular weight of the mixture:
\[ M = 74.25 \times 2 = 148.5 \, \text{g/mol} \]

4. **Calculate the Molecular Weight of PCl₅**:
The molecular weight of PCl₅ can be calculated as follows:
- Atomic weight of P = 31 g/mol
- Atomic weight of Cl = 35.5 g/mol
\[ M_{PCl_5} = 31 + (5 \times 35.5) = 31 + 177.5 = 208.5 \, \text{g/mol} \]

5. **Set Up the Dissociation Calculation**:
Let \( x \) be the number of moles of PCl₅ that dissociate. Initially, we have 1 mole of PCl₅.
- At equilibrium:
- Moles of PCl₅ = \( 1 - x \)
- Moles of PCl₃ = \( x \)
- Moles of Cl₂ = \( x \)

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

6. **Calculate the Average Molecular Weight at Equilibrium**:
The average molecular weight of the mixture can be expressed as:
\[ M_{\text{avg}} = \frac{(1 - x) \cdot 208.5 + x \cdot 137.5 + x \cdot 35.5}{1 + x} \]
Where:
- \( M_{PCl_3} = 137.5 \, \text{g/mol} \)
- \( M_{Cl_2} = 71 \, \text{g/mol} \)

Simplifying this gives:
\[ M_{\text{avg}} = \frac{(208.5 - 208.5x + 137.5x + 71x)}{1 + x} = \frac{208.5 - 208.5x + 208.5x}{1 + x} = \frac{208.5}{1 + x} \]

7. **Set Up the Equation**:
We know the average molecular weight from the vapour density:
\[ 148.5 = \frac{208.5}{1 + x} \]

Rearranging gives:
\[ 148.5(1 + x) = 208.5 \]
\[ 148.5 + 148.5x = 208.5 \]
\[ 148.5x = 208.5 - 148.5 = 60 \]
\[ x = \frac{60}{148.5} \approx 0.404 \]

8. **Calculate Percentage Dissociation**:
The percentage dissociation of PCl₅ is given by:
\[ \text{Percentage dissociation} = \left( \frac{x}{1} \right) \times 100 = 0.404 \times 100 \approx 40.4\% \]

### Final Answer:
The percentage dissociation of PCl₅ is approximately **40.4%**.