The molar conductance of NaCl, HCl and `CH_(3)COONa` at infinite dilution are 126.45, 426.16 and 91 `ohm^(-1) cm^(2) mol^(-1)` respectively. The molar conductance of `CH_(3)COOH` at infinite dilution is :

Options

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

Answer

To find the molar conductance of acetic acid (CH₃COOH) at infinite dilution, we can use the relationship between the molar conductance of the ions involved in the dissociation of the compounds.

### Step-by-Step Solution:

1. **Identify the given values**:
- Molar conductance of NaCl (Λ°(NaCl)) = 126.45 ohm⁻¹ cm² mol⁻¹
- Molar conductance of HCl (Λ°(HCl)) = 426.16 ohm⁻¹ cm² mol⁻¹
- Molar conductance of CH₃COONa (Λ°(CH₃COONa)) = 91 ohm⁻¹ cm² mol⁻¹

2. **Understand the dissociation of the compounds**:
- NaCl dissociates into Na⁺ and Cl⁻ ions.
- HCl dissociates into H⁺ and Cl⁻ ions.
- CH₃COONa dissociates into CH₃COO⁻ and Na⁺ ions.
- CH₃COOH is a weak acid that partially dissociates into CH₃COO⁻ and H⁺ ions.

3. **Use the relationship for molar conductance**:
The molar conductance of CH₃COOH at infinite dilution can be calculated using the formula:
\[ \Lambda°(CH₃COOH) = \Lambda°(CH₃COONa) + \Lambda°(HCl) - \Lambda°(NaCl) \]

4. **Substitute the values into the equation**:
\[ \Lambda°(CH₃COOH) = 91 + 426.16 - 126.45 \]

5. **Perform the calculations**:
- First, add 91 and 426.16:
\[ 91 + 426.16 = 517.16 \]
- Then, subtract 126.45 from 517.16:
\[ 517.16 - 126.45 = 390.71 \]

6. **Final result**:
The molar conductance of CH₃COOH at infinite dilution is:
\[ \Lambda°(CH₃COOH) = 390.71 \, \text{ohm}^{-1} \text{cm}^2 \text{mol}^{-1} \]