At infinite dilution, the equivalent conductances of `CH_(3)COOHa, HCl and CH_(3)COOH` are 91, 426 and 391 mho `cm^(2)eq^(-1)` respectively at `25^(@)C`. The equivalent conductance of NaCl at infinite dilution wil be:

Options

A. `126" ohm"^(-1) cm^(2) eq^(-1)`
B. `209" ohm"^(-1) cm^(2) eq^(-1)`
C. `391" ohm"^(-1) cm^(2) eq^(-1)`
D. `908" ohm"^(-1) cm^(2) eq^(-1)`

Answer

To find the equivalent conductance of NaCl at infinite dilution using the given values for CH₃COO⁻, HCl, and CH₃COOH, we can apply Kohlrausch's Law. This law states that the equivalent conductance of an electrolyte at infinite dilution is the sum of the conductances of its cations and anions.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Equivalent conductance of CH₃COO⁻ (sodium acetate) = 91 mho cm² eq⁻¹
- Equivalent conductance of HCl = 426 mho cm² eq⁻¹
- Equivalent conductance of CH₃COOH (acetic acid) = 391 mho cm² eq⁻¹

2. **Write the Dissociation Reactions:**
- For CH₃COONa: CH₃COONa → CH₃COO⁻ + Na⁺
- For HCl: HCl → H⁺ + Cl⁻
- For CH₃COOH: CH₃COOH ⇌ H⁺ + CH₃COO⁻

3. **Apply Kohlrausch's Law:**
- According to Kohlrausch's Law, the equivalent conductance of NaCl at infinite dilution (λ₀(NaCl)) can be expressed as:
\[ λ₀(NaCl) = λ₀(HCl) + λ₀(CH₃COO⁻) - λ₀(CH₃COOH) \]

4. **Substitute the Values:**
\[ λ₀(NaCl) = 426 \, \text{mho cm}^2 \text{eq}^{-1} + 91 \, \text{mho cm}^2 \text{eq}^{-1} - 391 \, \text{mho cm}^2 \text{eq}^{-1} \]

5. **Perform the Calculation:**
\[ λ₀(NaCl) = 426 + 91 - 391 \]
\[ λ₀(NaCl) = 126 \, \text{mho cm}^2 \text{eq}^{-1} \]

6. **Conclusion:**
The equivalent conductance of NaCl at infinite dilution is **126 mho cm² eq⁻¹**.