120 mm of pressure is developed at equilibrium when PCl5 ​ at 100 mm is subjected to decomposition. Then the percentage of dissociation of PCl5 is

Explanation

To solve the problem, we need to find the percentage of dissociation of PCl5 when it decomposes into PCl3 and Cl2. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
PCl5 decomposes according to the following reaction:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

### Step 2: Set Up Initial Conditions
Initially, we have:
- Pressure of PCl5 = 100 mm
- Pressure of PCl3 = 0 mm
- Pressure of Cl2 = 0 mm

### Step 3: Define Change in Pressure
Let \( P \) be the pressure of PCl5 that dissociates. At equilibrium, the pressures will be:
- Pressure of PCl5 = \( 100 - P \) mm
- Pressure of PCl3 = \( P \) mm
- Pressure of Cl2 = \( P \) mm

### Step 4: Write the Total Pressure at Equilibrium
According to the problem, the total pressure at equilibrium is 120 mm. Therefore, we can write:
\[ (100 - P) + P + P = 120 \]

### Step 5: Simplify the Equation
This simplifies to:
\[ 100 + P = 120 \]

### Step 6: Solve for P
Now, we can solve for \( P \):
\[ P = 120 - 100 \]
\[ P = 20 \text{ mm} \]

### Step 7: Calculate the Percentage of Dissociation
The percentage of dissociation of PCl5 can be calculated using the formula:
\[ \text{Percentage of dissociation} = \left( \frac{\text{Amount dissociated}}{\text{Initial amount}} \right) \times 100 \]

In this case, the amount dissociated is \( P = 20 \) mm, and the initial amount is 100 mm:
\[ \text{Percentage of dissociation} = \left( \frac{20}{100} \right) \times 100 = 20\% \]

### Final Answer
The percentage of dissociation of PCl5 is **20%**.

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