Molar conductivities `(Lambda_(m)^(@))` at infinite dilution of `NaCl, HCl` and `CH_(3)COONa` arc `126.4, 425.9` and `91.0 S cm^(2) mol^(-1)` respectively. `Lambda_(m)^(@)` for `CH_(3)COOH` will be

Options

A. `390.5 S cm^(2) mol^(-1)`
B. `425.5 S cm^(2) mol^(-1)`
C. `180.5 S cm^(2) mol^(-1)`
D. `290.8 S cm^(2) mol^(-1)`

Explanation

To find the molar conductivity at infinite dilution of acetic acid (CH₃COOH), we can use the Kohlrausch's Law, which states that the molar conductivity at infinite dilution is the sum of the molar conductivities of the individual ions.

### Step-by-Step Solution:

1. **Identify the Molar Conductivities**:
- Given values:
- Molar conductivity of NaCl (Λₘ(NaCl)) = 126.4 S cm² mol⁻¹
- Molar conductivity of HCl (Λₘ(HCl)) = 425.9 S cm² mol⁻¹
- Molar conductivity of sodium acetate (Λₘ(CH₃COONa)) = 91.0 S cm² mol⁻¹

2. **Write the Molar Conductivity Expressions**:
- For NaCl:
\[ Λₘ(NaCl) = Λ(Na^+) + Λ(Cl^-) \]
- For HCl:
\[ Λₘ(HCl) = Λ(H^+) + Λ(Cl^-) \]
- For sodium acetate:
\[ Λₘ(CH₃COONa) = Λ(Na^+) + Λ(CH₃COO^-) \]

3. **Set Up the Equation for Acetic Acid**:
- For acetic acid (CH₃COOH):
\[ Λₘ(CH₃COOH) = Λ(H^+) + Λ(CH₃COO^-) \]

4. **Relate the Conductivities**:
- We can express the molar conductivity of acetic acid in terms of the other three:
\[ Λ(CH₃COOH) = Λ(CH₃COO^-) + Λ(H^+) = Λ(CH₃COONa) + Λ(HCl) - Λ(NaCl) \]

5. **Substitute the Given Values**:
- Substitute the known values into the equation:
\[ Λ(CH₃COOH) = 91.0 + 425.9 - 126.4 \]

6. **Calculate**:
- Perform the calculation:
\[ Λ(CH₃COOH) = 91.0 + 425.9 - 126.4 = 390.5 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
The molar conductivity at infinite dilution of acetic acid (CH₃COOH) is **390.5 S cm² mol⁻¹**.

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