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

Answer

To find the molar conductance of acetic acid (CH₃COOH) at infinite dilution, we can use the following relationship based on the principle of additivity of molar conductances:

\[ \Lambda_m (\text{CH}_3\text{COOH}) = \Lambda_m (\text{CH}_3\text{COONa}) + \Lambda_m (\text{HCl}) - \Lambda_m (\text{NaCl}) \]

Where:
- \(\Lambda_m (\text{CH}_3\text{COOH})\) = Molar conductance of acetic acid
- \(\Lambda_m (\text{CH}_3\text{COONa})\) = Molar conductance of sodium acetate = 91 \(\Omega^{-1} \text{cm}^2 \text{mol}^{-1}\)
- \(\Lambda_m (\text{HCl})\) = Molar conductance of hydrochloric acid = 426.16 \(\Omega^{-1} \text{cm}^2 \text{mol}^{-1}\)
- \(\Lambda_m (\text{NaCl})\) = Molar conductance of sodium chloride = 126.45 \(\Omega^{-1} \text{cm}^2 \text{mol}^{-1}\)

### Step-by-step Calculation:

1. **Identify the given values**:
- \(\Lambda_m (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1}\)
- \(\Lambda_m (\text{HCl}) = 426.16 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1}\)
- \(\Lambda_m (\text{NaCl}) = 126.45 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1}\)

2. **Substitute the values into the equation**:
\[ \Lambda_m (\text{CH}_3\text{COOH}) = 91 + 426.16 - 126.45 \]

3. **Perform the addition and subtraction**:
- First, add \(91\) and \(426.16\):
\[ 91 + 426.16 = 517.16 \]
- Then, subtract \(126.45\) from \(517.16\):
\[ 517.16 - 126.45 = 390.71 \]

4. **Final result**:
\[ \Lambda_m (\text{CH}_3\text{COOH}) = 390.71 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \]

### Conclusion:
The molar conductance of acetic acid (CH₃COOH) at infinite dilution is \(390.71 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1}\).