`{:("Electrolyte",^^.^(oo)(Scm^(2)mol^(-1))),(KCl,149.9),(KNO_(3),145.0),(HCl,426.2),(NaOAc,91.0),(NaCl,126.5):}`
Calculate `^^_(HOAc)^(oo)` using appropriate molar conductance of the electrolytes listed above at infinite dilution in `H_(2)O` at `25^(@)C`

Options

A. ` 517. 2 `
B. `552.7`
C. ` 390.7 `
D. `217.5 `

Explanation

To calculate the molar conductance of acetic acid (HOAc) at infinite dilution (\( \Lambda^\infty_{HOAc} \)), we will use the molar conductance values of the given electrolytes at infinite dilution. The molar conductance of a weak electrolyte like acetic acid can be expressed in terms of its ions.

### Step-by-Step Solution:

1. **Identify the Molar Conductance of Given Electrolytes**:
- KCl: \( \Lambda^\infty_{KCl} = 149.9 \, \text{S cm}^2 \text{mol}^{-1} \)
- KNO₃: \( \Lambda^\infty_{KNO3} = 145.0 \, \text{S cm}^2 \text{mol}^{-1} \)
- HCl: \( \Lambda^\infty_{HCl} = 426.2 \, \text{S cm}^2 \text{mol}^{-1} \)
- NaOAc: \( \Lambda^\infty_{NaOAc} = 91.0 \, \text{S cm}^2 \text{mol}^{-1} \)
- NaCl: \( \Lambda^\infty_{NaCl} = 126.5 \, \text{S cm}^2 \text{mol}^{-1} \)

2. **Write the Molar Conductance Equations**:
- For KCl:
\[ \Lambda^\infty_{KCl} = \Lambda^\infty_{K^+} + \Lambda^\infty_{Cl^-} \]
- For KNO₃:
\[ \Lambda^\infty_{KNO3} = \Lambda^\infty_{K^+} + \Lambda^\infty_{NO_3^-} \]
- For HCl:
\[ \Lambda^\infty_{HCl} = \Lambda^\infty_{H^+} + \Lambda^\infty_{Cl^-} \]
- For NaOAc:
\[ \Lambda^\infty_{NaOAc} = \Lambda^\infty_{Na^+} + \Lambda^\infty_{OAc^-} \]
- For NaCl:
\[ \Lambda^\infty_{NaCl} = \Lambda^\infty_{Na^+} + \Lambda^\infty_{Cl^-} \]

3. **Express \( \Lambda^\infty_{HOAc} \)**:
- The molar conductance of acetic acid can be expressed as:
\[ \Lambda^\infty_{HOAc} = \Lambda^\infty_{H^+} + \Lambda^\infty_{OAc^-} \]

4. **Substitute Known Values**:
- From the equations, we can isolate \( \Lambda^\infty_{H^+} \) and \( \Lambda^\infty_{OAc^-} \):
- From HCl:
\[ \Lambda^\infty_{H^+} = \Lambda^\infty_{HCl} - \Lambda^\infty_{Cl^-} = 426.2 - \Lambda^\infty_{Cl^-} \]
- From NaCl:
\[ \Lambda^\infty_{Na^+} + \Lambda^\infty_{Cl^-} = 126.5 \]
- From NaOAc:
\[ \Lambda^\infty_{Na^+} + \Lambda^\infty_{OAc^-} = 91.0 \]

5. **Solve for \( \Lambda^\infty_{OAc^-} \)**:
- Now, we can substitute \( \Lambda^\infty_{Na^+} \) from the NaCl equation into the NaOAc equation:
\[ (126.5 - \Lambda^\infty_{Cl^-}) + \Lambda^\infty_{OAc^-} = 91.0 \]
- Rearranging gives:
\[ \Lambda^\infty_{OAc^-} = 91.0 - (126.5 - \Lambda^\infty_{Cl^-}) \]
- Simplifying yields:
\[ \Lambda^\infty_{OAc^-} = \Lambda^\infty_{Cl^-} - 35.5 \]

6. **Substituting Back**:
- Substitute \( \Lambda^\infty_{OAc^-} \) back into the equation for \( \Lambda^\infty_{HOAc} \):
\[ \Lambda^\infty_{HOAc} = \Lambda^\infty_{H^+} + \Lambda^\infty_{OAc^-} \]
- We can now express \( \Lambda^\infty_{HOAc} \) in terms of known values.

7. **Final Calculation**:
- After substituting and simplifying the equations, we find:
\[ \Lambda^\infty_{HOAc} = 390.7 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
\[ \Lambda^\infty_{HOAc} = 390.7 \, \text{S cm}^2 \text{mol}^{-1} \]