For the given equilibrium
`L_((g)) hArr M_((g))`
The `K_f =5xx10^(-4)` mole /litre /seconds and `k_b=3 xx 10^(-2)` litre/mole /seconds
the equilibrium concentration of M is

Options

A. 0.13 M
B. 0.3 M
C. 0.8 M
D. can't be calculated

Answer

To find the equilibrium concentration of M in the reaction \( L_{(g)} \rightleftharpoons M_{(g)} \), we will use the relationship between the rate constants of the forward and backward reactions.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Forward rate constant, \( K_f = 5 \times 10^{-4} \) mole/litre/seconds
- Backward rate constant, \( K_b = 3 \times 10^{-2} \) litre/mole/seconds

2. **Use the Relationship Between \( K_f \) and \( K_b \)**:
- The equilibrium constant \( K_c \) can be determined using the formula:
\[ K_c = \frac{K_f}{K_b} \]

3. **Substituting the Values**:
- Substitute the values of \( K_f \) and \( K_b \) into the equation:
\[ K_c = \frac{5 \times 10^{-4}}{3 \times 10^{-2}} \]

4. **Calculate \( K_c \)**:
- Performing the division:
\[ K_c = \frac{5}{3} \times 10^{-4 + 2} = \frac{5}{3} \times 10^{-2} \]
- Now, calculate \( \frac{5}{3} \):
\[ \frac{5}{3} \approx 1.6667 \]
- Therefore,
\[ K_c \approx 1.6667 \times 10^{-2} \approx 0.01667 \]

5. **Final Result**:
- The equilibrium constant \( K_c \) is approximately \( 0.01667 \).

### Conclusion:
The equilibrium concentration of M is represented by the equilibrium constant \( K_c \), which we calculated to be approximately \( 0.01667 \).