What is the vapour density of mixture of `PCL_(5)` at `250^(@)C` when it has dissociated to the extent of `80%` ?

Options

A. 58
B. `41.7`
C. `52.25`
D. `83.6`

Answer

To find the vapor density of a mixture of PCl₅ at 250°C when it has dissociated to the extent of 80%, we can follow these steps:

### Step 1: Understand the dissociation of PCl₅
PCl₅ dissociates into PCl₃ and Cl₂ according to the reaction:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

### Step 2: Define the degree of dissociation
Let the degree of dissociation (α) be 80%, which can be expressed as:
\[ \alpha = 0.8 \]

### Step 3: Determine the initial and equilibrium moles
Assume we start with 1 mole of PCl₅. At equilibrium:
- Moles of PCl₅ remaining = \( 1 - \alpha = 1 - 0.8 = 0.2 \)
- Moles of PCl₃ formed = \( \alpha = 0.8 \)
- Moles of Cl₂ formed = \( \alpha = 0.8 \)

Thus, the total number of moles at equilibrium is:
\[ \text{Total moles} = \text{Moles of PCl}_5 + \text{Moles of PCl}_3 + \text{Moles of Cl}_2 \]
\[ = (1 - 0.8) + 0.8 + 0.8 = 0.2 + 0.8 + 0.8 = 1.8 \]

### Step 4: Calculate the initial vapor density
The molecular mass of PCl₅ is given as 208.5 g/mol. The initial vapor density (VD_initial) can be calculated using the formula:
\[ \text{VD}_{\text{initial}} = \frac{\text{Molecular weight}}{2} \]
\[ \text{VD}_{\text{initial}} = \frac{208.5}{2} = 104.25 \]

### Step 5: Relate vapor density to the number of moles
The vapor density at equilibrium (VD_equilibrium) can be calculated using the relationship:
\[ \text{VD}_{\text{equilibrium}} = \text{VD}_{\text{initial}} \times \frac{\text{Number of moles}_{\text{initial}}}{\text{Number of moles}_{\text{equilibrium}}} \]
Where:
- Number of moles_initial = 1 (initially we had 1 mole of PCl₅)
- Number of moles_equilibrium = 1.8 (calculated in Step 3)

Substituting the values:
\[ \text{VD}_{\text{equilibrium}} = 104.25 \times \frac{1}{1.8} \]
\[ \text{VD}_{\text{equilibrium}} = 104.25 \times 0.5556 \approx 57.91 \]

### Step 6: Final answer
The vapor density of the mixture of PCl₅ at 250°C when it has dissociated to the extent of 80% is approximately:
\[ \text{VD}_{\text{equilibrium}} \approx 58 \]