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 solve the problem of finding the vapor density of a mixture of \( PCl_5 \) at \( 250^\circ C \) when it has dissociated to the extent of \( 80\% \), we can follow these steps:

### Step 1: Understanding the Dissociation of \( PCl_5 \)
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + \frac{1}{2} Cl_2 \]
When \( PCl_5 \) dissociates, it produces \( PCl_3 \) and \( Cl_2 \).

### Step 2: Initial and Equilibrium Moles
Let's assume we start with 1 mole of \( PCl_5 \).

- Initial moles of \( PCl_5 \) = 1
- Moles of \( PCl_3 \) and \( Cl_2 \) at the start = 0

At equilibrium, if \( 80\% \) of \( PCl_5 \) dissociates, then:
- Moles of \( PCl_5 \) remaining = \( 1 - 0.8 = 0.2 \)
- Moles of \( PCl_3 \) produced = \( 0.8 \)
- Moles of \( Cl_2 \) produced = \( 0.4 \) (since \( \frac{1}{2} \) mole of \( Cl_2 \) is produced for every mole of \( PCl_5 \) that dissociates)

### Step 3: Total Moles at Equilibrium
Now, we can calculate the total moles at equilibrium:
\[ \text{Total moles} = \text{Moles of } PCl_5 + \text{Moles of } PCl_3 + \text{Moles of } Cl_2 \]
\[ \text{Total moles} = 0.2 + 0.8 + 0.4 = 1.4 \]

### Step 4: Molar Mass of \( PCl_5 \)
Next, we need to calculate the molar mass of \( PCl_5 \):
- Molar mass of \( P \) = 31 g/mol
- Molar mass of \( Cl \) = 35.5 g/mol

Thus, the molar mass of \( PCl_5 \):
\[ \text{Molar mass of } PCl_5 = 31 + (5 \times 35.5) = 31 + 177.5 = 208.5 \text{ g/mol} \]

### Step 5: Initial Vapor Density
The initial vapor density (\( D_i \)) can be calculated using the formula:
\[ D = \frac{\text{Molar mass}}{2} \]
So,
\[ D_i = \frac{208.5}{2} = 104.25 \]

### Step 6: Vapor Density at Equilibrium
The vapor density at equilibrium (\( D_e \)) can be calculated using the formula:
\[ D_e = D_i \times \frac{\text{Molar mass at equilibrium}}{\text{Initial molar mass}} \]
The molar mass at equilibrium can be calculated as:
\[ \text{Molar mass at equilibrium} = \frac{(0.2 \times 208.5) + (0.8 \times 137.5) + (0.4 \times 35.5)}{1.4} \]
Where:
- Molar mass of \( PCl_3 \) = \( 31 + (3 \times 35.5) = 31 + 106.5 = 137.5 \) g/mol
- Molar mass of \( Cl_2 \) = \( 2 \times 35.5 = 71 \) g/mol

Calculating the total molar mass at equilibrium:
\[ \text{Total molar mass} = (0.2 \times 208.5) + (0.8 \times 137.5) + (0.4 \times 71) \]
\[ = 41.7 + 110 + 28.4 = 180.1 \]
Now, substituting this back into the vapor density formula:
\[ D_e = 104.25 \times \frac{180.1}{208.5} \]
Calculating \( D_e \):
\[ D_e \approx 104.25 \times 0.864 \approx 90.0 \]

### Final Answer
The vapor density of the mixture of \( PCl_5 \) at \( 250^\circ C \) when it has dissociated to the extent of \( 80\% \) is approximately **90.0 g/L**.