The vapour density of `PCl_(5)` at `43K` is is found to be `70.2`. Find the degree of dissociation of `PCl_(5)` at this temperature.

Explanation

To find the degree of dissociation (α) of \( PCl_5 \) at 43 K, we can follow these steps:

### Step 1: Understand the Reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5 \rightleftharpoons PCl_3 + Cl_2 \]
From this reaction, we can see that 1 mole of \( PCl_5 \) produces 1 mole of \( PCl_3 \) and 1 mole of \( Cl_2 \), resulting in a total of 2 moles of products.

### Step 2: Determine the Initial and Equilibrium Vapor Densities
We are given the vapor density of \( PCl_5 \) at equilibrium (denoted as \( d \)) as 70.2. We need to calculate the initial vapor density (denoted as \( D \)).

### Step 3: Calculate the Molecular Mass of \( PCl_5 \)
The molecular mass of \( PCl_5 \) can be calculated as follows:
- Atomic mass of Phosphorus (P) = 31 g/mol
- Atomic mass of Chlorine (Cl) = 35.5 g/mol

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

### Step 4: Calculate the Initial Vapor Density \( D \)
The vapor density \( D \) is related to the molecular mass by the formula:
\[ D = \frac{M}{2} \]
Substituting the molecular mass of \( PCl_5 \):
\[ D = \frac{208.5}{2} = 104.25 \]

### Step 5: Apply the Formula for Degree of Dissociation
The formula for degree of dissociation \( \alpha \) is given by:
\[ \alpha = \frac{D - d}{(n - 1) \cdot d} \]
Where:
- \( D = 104.25 \)
- \( d = 70.2 \)
- \( n = 2 \) (since 1 mole of \( PCl_5 \) produces 2 moles of products)

### Step 6: Substitute the Values into the Formula
Now substituting the values into the formula:
\[ \alpha = \frac{104.25 - 70.2}{(2 - 1) \cdot 70.2} \]
\[ \alpha = \frac{34.05}{1 \cdot 70.2} = \frac{34.05}{70.2} \]

### Step 7: Calculate the Degree of Dissociation
Now calculating the value:
\[ \alpha \approx 0.485 \]

### Step 8: Convert to Percentage
To express the degree of dissociation as a percentage:
\[ \alpha \times 100 = 0.485 \times 100 \approx 48.5\% \]

### Final Answer
The degree of dissociation of \( PCl_5 \) at 43 K is approximately **48.5%**.

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