How much `PCl_(5)` must be added to a one litre vessel at `250^(@)C` in order to obtain a `35` concentration of `0.1` mol of `Cl_(2)`? `K_(c)` for `PCl_(5) hArr PCl_(3)+Cl_(2)` is `0.0414 mol L^(-1)`

Explanation

To solve the problem, we need to determine how much PCl₅ must be added to a 1-liter vessel at 250°C to achieve a concentration of 0.1 moles of Cl₂, given that the equilibrium constant (Kc) for the reaction is 0.0414 mol/L.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The dissociation of PCl₅ can be represented as:
\[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \]

2. **Define Initial Concentrations:**
Let \( A \) be the initial concentration of PCl₅ (in moles). At the start (t=0), we have:
- [PCl₅] = A
- [PCl₃] = 0
- [Cl₂] = 0

3. **Define Change in Concentrations:**
Let \( x \) be the amount of PCl₅ that dissociates. At equilibrium (t=equilibrium), the concentrations will be:
- [PCl₅] = \( A - x \)
- [PCl₃] = \( x \)
- [Cl₂] = \( x \)

4. **Use the Given Information:**
We know that at equilibrium, the concentration of Cl₂ is 0.1 mol/L. Therefore, we can set \( x = 0.1 \).

5. **Set Up the Equilibrium Expression:**
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \]
Substituting the equilibrium concentrations into the expression gives:
\[ K_c = \frac{x \cdot x}{A - x} = \frac{x^2}{A - x} \]
Given that \( K_c = 0.0414 \) and \( x = 0.1 \):
\[ 0.0414 = \frac{(0.1)^2}{A - 0.1} \]

6. **Solve for A:**
Rearranging the equation:
\[ 0.0414(A - 0.1) = 0.01 \]
Expanding this gives:
\[ 0.0414A - 0.00414 = 0.01 \]
Adding 0.00414 to both sides:
\[ 0.0414A = 0.01414 \]
Dividing both sides by 0.0414:
\[ A = \frac{0.01414}{0.0414} \approx 0.3415 \text{ moles} \]

7. **Conclusion:**
Therefore, the amount of PCl₅ that must be added to the vessel is approximately **0.3415 moles**.