1 mol of A and 0.5 mole of B were enclosed in a three litres vessel. The following equilibrium was established under suitable conditions.
`A+2B hArrC`
At equilibrium, the amount of B was found to be 0.3 mol. The equilibrium constant `K_C` at the experimental temperature will be

Options

A. `11.1`
B. `1.11`
C. `0.01 `
D. `2.50`

Explanation

To solve the problem, we need to find the equilibrium constant \( K_C \) for the reaction:

\[ A + 2B \rightleftharpoons C \]

### Step-by-Step Solution:

1. **Initial Moles**:
- Moles of A = 1 mol
- Moles of B = 0.5 mol
- Moles of C = 0 mol (since it is not mentioned, we assume it starts from zero)

2. **Change in Moles**:
- Let \( x \) be the amount of A that reacts to reach equilibrium.
- From the stoichiometry of the reaction, for every 1 mole of A that reacts, 2 moles of B react and 1 mole of C is produced.
- Therefore, at equilibrium:
- Moles of A = \( 1 - x \)
- Moles of B = \( 0.5 - 2x \)
- Moles of C = \( x \)

3. **Equilibrium Condition**:
- We know that at equilibrium, the amount of B is 0.3 mol. Thus, we can set up the equation:
\[ 0.5 - 2x = 0.3 \]

4. **Solving for \( x \)**:
- Rearranging the equation gives:
\[ 2x = 0.5 - 0.3 \]
\[ 2x = 0.2 \]
\[ x = 0.1 \]

5. **Calculating Moles at Equilibrium**:
- Now we can find the moles of A, B, and C at equilibrium:
- Moles of A = \( 1 - x = 1 - 0.1 = 0.9 \) mol
- Moles of B = \( 0.5 - 2x = 0.5 - 0.2 = 0.3 \) mol (as given)
- Moles of C = \( x = 0.1 \) mol

6. **Calculating Concentrations**:
- The volume of the vessel is 3 L, so we can calculate the concentrations:
- Concentration of A = \( \frac{0.9 \, \text{mol}}{3 \, \text{L}} = 0.3 \, \text{M} \)
- Concentration of B = \( \frac{0.3 \, \text{mol}}{3 \, \text{L}} = 0.1 \, \text{M} \)
- Concentration of C = \( \frac{0.1 \, \text{mol}}{3 \, \text{L}} = \frac{1}{30} \, \text{M} \approx 0.0333 \, \text{M} \)

7. **Calculating the Equilibrium Constant \( K_C \)**:
- The equilibrium constant \( K_C \) is given by the expression:
\[ K_C = \frac{[C]}{[A][B]^2} \]
- Substituting the concentrations:
\[ K_C = \frac{0.0333}{(0.3)(0.1)^2} \]
\[ K_C = \frac{0.0333}{(0.3)(0.01)} = \frac{0.0333}{0.003} \approx 11.1 \]

### Final Answer:
The equilibrium constant \( K_C \) at the experimental temperature is approximately **11.1**.