Equivalent conductance of NaCl, HCl and `C_(2)H_(5)COONa` at infinite dilution are 126.45, 426.16 and 91 `"ohm"^(-1) "cm"^(2)` respectively. The equivalent conductance of `C_(2)H_(5)COOH` is

Options

A. `201.28"ohm"^(-1)//cm^(-2)`
B. `390.71//"ohm"^(-1)//cm^(-2)`
C. `698.28//"ohm cm"^(2)`
D. `540.48//"ohm cm"^(2)`

Explanation

To find the equivalent conductance of acetic acid (C₂H₄O₂) at infinite dilution using the provided data, we can apply Kohlrausch's Law. This law states that the equivalent conductance of a weak electrolyte can be determined by the sum of the equivalent conductances of its ions at infinite dilution.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Equivalent conductance of NaCl (Λ(NaCl)) = 126.45 ohm⁻¹ cm²
- Equivalent conductance of HCl (Λ(HCl)) = 426.16 ohm⁻¹ cm²
- Equivalent conductance of sodium propanoate (Λ(C₂H₅COONa)) = 91 ohm⁻¹ cm²

2. **Write the Reaction for Acetic Acid**:
- The dissociation of acetic acid can be represented as:
\[ C_2H_5COOH \rightleftharpoons C_2H_5COO^- + H^+ \]
- Here, C₂H₅COO⁻ is the propanoate ion and H⁺ is the hydrogen ion.

3. **Apply Kohlrausch's Law**:
- According to Kohlrausch's Law, the equivalent conductance of acetic acid (Λ(C₂H₅COOH)) can be calculated as:
\[ Λ(C_2H_5COOH) = Λ(HCl) + Λ(C_2H_5COONa) - Λ(NaCl) \]

4. **Substitute the Values**:
- Now, substitute the given values into the equation:
\[ Λ(C_2H_5COOH) = 426.16 + 91 - 126.45 \]

5. **Perform the Calculation**:
- Calculate the equivalent conductance:
\[ Λ(C_2H_5COOH) = 426.16 + 91 - 126.45 = 390.71 \, \text{ohm}^{-1} \text{cm}^2 \]

6. **Conclusion**:
- The equivalent conductance of acetic acid (C₂H₄O₂) at infinite dilution is **390.71 ohm⁻¹ cm²**.