At `473 K`, partially dissociated vapours of `PCl_(5)` are `62` times as heavy as `H_(2)`. Calculate the degree of dissociation of `PCl_(5)`.

Explanation

To solve the problem of calculating the degree of dissociation of \( PCl_5 \) at \( 473 K \), we can follow these steps:

### Step 1: Write the dissociation reaction
The dissociation of \( PCl_5 \) can be represented as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Define the initial and equilibrium conditions
Assume we start with 1 mole of \( PCl_5 \). Let \( \alpha \) be the degree of dissociation. At equilibrium:
- Moles of \( PCl_5 \) = \( 1 - \alpha \)
- Moles of \( PCl_3 \) = \( \alpha \)
- Moles of \( Cl_2 \) = \( \alpha \)

### Step 3: Calculate the total number of moles at equilibrium
The total moles at equilibrium will be:
\[ \text{Total moles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

### Step 4: Calculate the theoretical molar mass (\( M_F \))
The molar mass of \( PCl_5 \) is calculated as follows:
- Molar mass of \( P \) = 31 g/mol
- Molar mass of \( Cl \) = 35.5 g/mol
- Therefore, \( M_F = 31 + 5 \times 35.5 = 31 + 177.5 = 208.5 \, \text{g/mol} \)

### Step 5: Calculate the observed molar mass (\( M_O \))
The problem states that the vapor of \( PCl_5 \) is 62 times as heavy as \( H_2 \). The molar mass of \( H_2 \) is 2 g/mol. Thus:
\[ M_O = 62 \times 2 = 124 \, \text{g/mol} \]

### Step 6: Use the formula for degree of dissociation
The degree of dissociation (\( \alpha \)) can be calculated using the formula:
\[ \alpha = \frac{M_F - M_O}{M_F} \]
Substituting the values we have:
\[ \alpha = \frac{208.5 - 124}{208.5} \]
\[ \alpha = \frac{84.5}{208.5} \approx 0.405 \]

### Step 7: Convert to percentage
To express the degree of dissociation as a percentage:
\[ \alpha \times 100 = 0.405 \times 100 \approx 40.5\% \]

### Final Answer
The degree of dissociation of \( PCl_5 \) is approximately \( 40.5\% \).

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