Calculate molar conductivity at infinite dilution of `CH_3COOH` if molar conductivity at infinite dilution of `CH_3COONa, HCI` and `MaCI` are `91.6, 425.0` and ` 128.1 S cm^2 "mol"^(-1)`.

Options

A. ` 390 .5 S cm^2 "mol"^(-1)`
B. ` 388.5 S cm^2 "mol"^(-1)`
C. ` 490.5 S cm^2 "mol"^(-1)`
D. None of these

Explanation

To calculate the molar conductivity at infinite dilution of acetic acid (CH₃COOH), we can use the molar conductivities of the salts and acids provided. The molar conductivities at infinite dilution for the substances are:

- CH₃COONa (sodium acetate): λ(CH₃COONa) = 91.6 S cm² mol⁻¹
- HCl (hydrochloric acid): λ(HCl) = 425.0 S cm² mol⁻¹
- NaCl (sodium chloride): λ(NaCl) = 128.1 S cm² mol⁻¹

### Step-by-Step Solution:

1. **Write the equations for the molar conductivities of the compounds:**
- For sodium acetate (CH₃COONa):
\[ \lambda_{CH₃COONa} = \lambda_{CH₃COO^-} + \lambda_{Na^+} \]
- For hydrochloric acid (HCl):
\[ \lambda_{HCl} = \lambda_{H^+} + \lambda_{Cl^-} \]
- For sodium chloride (NaCl):
\[ \lambda_{NaCl} = \lambda_{Na^+} + \lambda_{Cl^-} \]

2. **Express the molar conductivity of acetic acid (CH₃COOH):**
- The dissociation of acetic acid can be represented as:
\[ CH₃COOH \rightleftharpoons CH₃COO^- + H^+ \]
- Therefore, the molar conductivity at infinite dilution for acetic acid is:
\[ \lambda_{CH₃COOH} = \lambda_{CH₃COO^-} + \lambda_{H^+} \]

3. **Substitute the values to find λ(CH₃COO⁻) and λ(H⁺):**
- From the equations for NaCl and HCl, we can express λ(Na⁺) and λ(Cl⁻):
\[ \lambda_{Na^+} = \lambda_{NaCl} - \lambda_{Cl^-} \]
- Rearranging gives:
\[ \lambda_{Cl^-} = \lambda_{NaCl} - \lambda_{Na^+} \]
- Substitute λ(NaCl) = 128.1 S cm² mol⁻¹ into the equation:
\[ \lambda_{Cl^-} = 128.1 - \lambda_{Na^+} \]

4. **Combine the equations:**
- We can combine the equations for sodium acetate and hydrochloric acid:
\[ \lambda_{CH₃COO^-} = \lambda_{CH₃COONa} - \lambda_{Na^+} \]
- Substitute this into the equation for λ(CH₃COOH):
\[ \lambda_{CH₃COOH} = (\lambda_{CH₃COONa} - \lambda_{Na^+}) + \lambda_{H^+} \]

5. **Substituting the known values:**
- We have:
\[ \lambda_{CH₃COOH} = 91.6 + 425.0 - 128.1 \]
- Calculate:
\[ \lambda_{CH₃COOH} = 388.5 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
The molar conductivity at infinite dilution of CH₃COOH is **388.5 S cm² mol⁻¹**.