The vapour density of `Pcl_(5)` is `104.16` but when heated to `230^(@)C`, its vapour density is reduced to `62`. The degree of dissociation of `PCl_(5)` at `230^(@)C` is …………

Options

A. `6.8%`
B. `68%`
C. `46%`
D. `64%`

Answer

To find the degree of dissociation of \( PCl_5 \) at \( 230^\circ C \), we can follow these steps:

### Step 1: Understand the Reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]
Initially, we have 1 mole of \( PCl_5 \) and no products.

### Step 2: Define the Degree of Dissociation
Let \( \alpha \) be the degree of dissociation of \( PCl_5 \). At equilibrium:
- The moles of \( PCl_5 \) will be \( 1 - \alpha \)
- The moles of \( PCl_3 \) will be \( \alpha \)
- The moles of \( Cl_2 \) will be \( \alpha \)

### Step 3: Calculate Total Moles at Equilibrium
The total moles at equilibrium can be expressed as:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Relate Vapor Density to Moles
Vapor density is inversely proportional to the number of moles. Thus, we can set up the following relationship:
\[ \frac{\text{Initial Vapor Density}}{\text{Final Vapor Density}} = \frac{\text{Initial Moles}}{\text{Final Moles}} \]
Substituting the values we have:
\[ \frac{104.16}{62} = \frac{1}{1 + \alpha} \]

### Step 5: Solve for \( \alpha \)
Cross-multiplying gives:
\[ 104.16 \times (1 + \alpha) = 62 \]
Expanding this:
\[ 104.16 + 104.16\alpha = 62 \]
Rearranging to isolate \( \alpha \):
\[ 104.16\alpha = 62 - 104.16 \]
\[ 104.16\alpha = -42.16 \]
\[ \alpha = \frac{-42.16}{104.16} \]
Calculating \( \alpha \):
\[ \alpha = 0.68 \]

### Step 6: Convert to Percentage
To find the degree of dissociation in percentage:
\[ \text{Degree of dissociation} = \alpha \times 100 = 0.68 \times 100 = 68\% \]

### Final Answer
The degree of dissociation of \( PCl_5 \) at \( 230^\circ C \) is **68%**.

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