The molar conductivity of 0.05 M solution of weak acid is `16.6Omega^(-1)cm^(-2)mol^(-1).` Its molar conductivity at infinite dilution is `390.7Omega^(-1)cm^(-2)mol^(-1)`. The degree of dissociation of weak acid is

Options

A. 0.24
B. 0.42
C. 0.042
D. 0.024

Answer

To find the degree of dissociation (α) of the weak acid, we can use the formula that relates molar conductivity at a given concentration (Λ) and molar conductivity at infinite dilution (Λ°):

\[ \alpha = \frac{\Lambda}{\Lambda^\circ} \]

Where:
- \(\Lambda\) = Molar conductivity of the solution at a given concentration (in this case, 0.05 M)
- \(\Lambda^\circ\) = Molar conductivity at infinite dilution

Given:
- \(\Lambda = 16.6 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)
- \(\Lambda^\circ = 390.7 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)

### Step-by-Step Solution:

1. **Identify the values**:
- Molar conductivity of the weak acid solution (Λ) = 16.6 \( \Omega^{-1} cm^2 mol^{-1} \)
- Molar conductivity at infinite dilution (Λ°) = 390.7 \( \Omega^{-1} cm^2 mol^{-1} \)

2. **Substitute the values into the formula**:
\[ \alpha = \frac{16.6}{390.7} \]

3. **Calculate the degree of dissociation**:
\[ \alpha = \frac{16.6}{390.7} \approx 0.0425 \]

4. **Express α as a percentage (optional)**:
\[ \alpha \approx 0.0425 \times 100 = 4.25\% \]

### Final Answer:
The degree of dissociation (α) of the weak acid is approximately 0.0425 or 4.25%.