`k=4.95xx10^(-5) S cm^(-1)` for a `0.001 M `solution. The reciprocal of the degree of dissociation of acetic acid, if `wedge_(m)^(@)` for acetic acid is `400S cm^(-2) mol^(-1)` is `:`

Options

A. 7
B. 8
C. 9
D. 10

Explanation

To solve the problem, we need to find the reciprocal of the degree of dissociation (α) of acetic acid given the conductivity (k) and the molar conductivity at infinite dilution (λ_m^0).

### Step-by-Step Solution:

1. **Given Data**:
- Conductivity (k) = \(4.95 \times 10^{-5} \, \text{S cm}^{-1}\)
- Concentration (C) = \(0.001 \, \text{M}\)
- Molar conductivity at infinite dilution (λ_m^0) = \(400 \, \text{S cm}^2 \text{mol}^{-1}\)

2. **Calculate Molar Conductivity (λ_m)**:
The molar conductivity (λ_m) at a given concentration can be calculated using the formula:
\[ \lambda_m = \frac{1000 \cdot k}{C} \]
Substituting the values:
\[ \lambda_m = \frac{1000 \cdot (4.95 \times 10^{-5})}{0.001} \]
\[ \lambda_m = \frac{4.95 \times 10^{-2}}{0.001} = 4.95 \times 10^{1} = 49.5 \, \text{S cm}^2 \text{mol}^{-1} \]

3. **Calculate Degree of Dissociation (α)**:
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\lambda_m}{\lambda_m^0} \]
Substituting the values:
\[ \alpha = \frac{49.5}{400} \]

4. **Calculate the Reciprocal of Degree of Dissociation (1/α)**:
To find the reciprocal of the degree of dissociation:
\[ \frac{1}{\alpha} = \frac{\lambda_m^0}{\lambda_m} = \frac{400}{49.5} \]
Performing the division:
\[ \frac{1}{\alpha} \approx 8.08 \]

5. **Final Answer**:
Rounding off, we find that the reciprocal of the degree of dissociation is approximately:
\[ \frac{1}{\alpha} \approx 8 \]

### Conclusion:
The reciprocal of the degree of dissociation of acetic acid is **8**.