`0.1 mol` of `PCl_(5)` is vaporised in a litre vessel at `260^(@)C`. Calculate the concentration of `Cl_(2)` at equilibrium, if the equilibrium constant for the dissociation of `PCl_(5)` is `0.0414`.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Write the balanced chemical equation for the dissociation of PCl5.
The dissociation of phosphorus pentachloride (PCl5) can be represented as:
\[ \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \]

### Step 2: Set up the initial concentrations.
Given that we have 0.1 moles of PCl5 in a 1-liter vessel, the initial concentration of PCl5 is:
\[ [\text{PCl}_5] = 0.1 \, \text{mol/L} \]
At the start, the concentrations of PCl3 and Cl2 are both 0.

### Step 3: Define the change in concentration at equilibrium.
Let \( x \) be the amount of PCl5 that dissociates at equilibrium. Therefore, at equilibrium:
- The concentration of PCl5 will be \( 0.1 - x \)
- The concentration of PCl3 will be \( x \)
- The concentration of Cl2 will also be \( x \)

### Step 4: Write the expression for the equilibrium constant (Kc).
The equilibrium constant expression for the reaction is given by:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]
Substituting the equilibrium concentrations into the expression, we get:
\[ K_c = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]

### Step 5: Substitute the value of Kc and solve for x.
We know that \( K_c = 0.0414 \). Therefore, we can set up the equation:
\[ \frac{x^2}{0.1 - x} = 0.0414 \]
Cross-multiplying gives:
\[ x^2 = 0.0414(0.1 - x) \]
\[ x^2 = 0.00414 - 0.0414x \]
Rearranging this into standard quadratic form:
\[ x^2 + 0.0414x - 0.00414 = 0 \]

### Step 6: Solve the quadratic equation.
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 0.0414 \), and \( c = -0.00414 \):
1. Calculate the discriminant:
\[ b^2 - 4ac = (0.0414)^2 - 4(1)(-0.00414) \]
\[ = 0.00171396 + 0.01656 = 0.01827396 \]
2. Calculate \( x \):
\[ x = \frac{-0.0414 \pm \sqrt{0.01827396}}{2} \]
\[ = \frac{-0.0414 \pm 0.135 \, (approximately)}{2} \]
This gives two potential solutions for \( x \):
- \( x \approx 0.0468 \) (valid, since it's positive)
- \( x \approx -0.0884 \) (not valid, since concentration cannot be negative)

### Step 7: Determine the concentration of Cl2 at equilibrium.
Since \( [\text{Cl}_2] = x \), we find:
\[ [\text{Cl}_2] = 0.0468 \, \text{mol/L} \]

### Final Answer:
The concentration of Cl2 at equilibrium is:
\[ [\text{Cl}_2] = 0.0468 \, \text{mol/L} \]