The equivalent conductance of 1M benzoic acid is `12.8 ohm^(-1) cm^(2)` and if the conductance of benzoate ion and `H^+` ion at infinite dilution are `42 ohm^(-1) cm^(2)` and `288.42 ohm^(-1) cm^(2)` respectively . Then its degree of dissociation is

Options

A. `0.039%`
B. `3.9%`
C. `0.35%`
D. `39%`

Answer

To find the degree of dissociation (α) of benzoic acid, we can use the information provided about its equivalent conductance and the conductance of its ions at infinite dilution.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Equivalent conductance of 1M benzoic acid (C6H5COOH): \( \Lambda_{C6H5COOH} = 12.8 \, \Omega^{-1} \, cm^2 \)
- Conductance of benzoate ion (C6H5COO⁻) at infinite dilution: \( \Lambda_{C6H5COO^-} = 42 \, \Omega^{-1} \, cm^2 \)
- Conductance of H⁺ ion at infinite dilution: \( \Lambda_{H^+} = 288.42 \, \Omega^{-1} \, cm^2 \)

2. **Calculate the Limiting Equivalent Conductance of Benzoic Acid:**
According to Kohlrausch's law, the limiting equivalent conductance of a weak electrolyte is the sum of the conductances of its ions at infinite dilution:
\[ \Lambda_{C6H5COOH}^{\infty} = \Lambda_{C6H5COO^-} + \Lambda_{H^+} \]
Substituting the values:
\[ \Lambda_{C6H5COOH}^{\infty} = 42 + 288.42 = 330.42 \, \Omega^{-1} \, cm^2 \]

3. **Calculate the Degree of Dissociation (α):**
The degree of dissociation is given by the formula:
\[ \alpha = \frac{\Lambda_{C6H5COOH}}{\Lambda_{C6H5COOH}^{\infty}} \]
Substituting the values:
\[ \alpha = \frac{12.8}{330.42} \]
Calculating this gives:
\[ \alpha \approx 0.0387 \]

4. **Convert Degree of Dissociation to Percentage:**
To express the degree of dissociation as a percentage, multiply by 100:
\[ \text{Percentage of dissociation} = \alpha \times 100 = 0.0387 \times 100 \approx 3.87\% \]

### Final Answer:
The degree of dissociation of benzoic acid is approximately **3.87%**.