Limiting molar conductivity of `CH_(3)COOH (i.e.`^^_(m(CH_(3)COOH))^(0)` is equal to :-

Options

A. `^^_(m(CH_(3)COOH))^(@)+^^_(m(CH_(3)COONa))^(@)+^^_(m(NaOH))^(@)`
B. `^^_(m(CH_(3)COONa))^(@)+^^_(m(HCI))^(@)+^^_(m(NaCI))^(@)`
C. `^^_(m(CH_(3)COONa))^(@)+^^_(m(NaCI))^(@)+^^_(m(NaOH))^(@)`
D. `^^_(m(NaOH))^(@)+^^_(m(NaCI))^(@)+^^_(m(CH_(3)COONa))^(@)`

Explanation

To find the limiting molar conductivity of acetic acid (CH₃COOH), we can use Kohlrausch's law, which states that the limiting molar conductivity of a compound can be expressed in terms of the limiting molar conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Identify the ions in acetic acid**:
Acetic acid dissociates into acetate ions (CH₃COO⁻) and hydrogen ions (H⁺) in solution:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

2. **Use Kohlrausch's law**:
According to Kohlrausch's law, the limiting molar conductivity (Λ₀) of acetic acid can be expressed as:
\[ \Lambda_0(\text{CH}_3\text{COOH}) = \Lambda_0(\text{CH}_3\text{COO}^-) + \Lambda_0(\text{H}^+) \]
where:
- \(\Lambda_0(\text{CH}_3\text{COO}^-)\) is the limiting molar conductivity of the acetate ion.
- \(\Lambda_0(\text{H}^+)\) is the limiting molar conductivity of the hydrogen ion.

3. **Account for the ions from sodium acetate and HCl**:
To find the limiting molar conductivity of acetic acid, we can use the conductivities of sodium acetate (CH₃COONa) and hydrochloric acid (HCl), and subtract the conductivity of sodium chloride (NaCl) to eliminate the extra ions:
\[ \Lambda_0(\text{CH}_3\text{COOH}) = \Lambda_0(\text{CH}_3\text{COONa}) + \Lambda_0(\text{HCl}) - \Lambda_0(\text{NaCl}) \]

4. **Identify the components**:
- \(\Lambda_0(\text{CH}_3\text{COONa})\) provides the acetate ion (CH₃COO⁻) and sodium ion (Na⁺).
- \(\Lambda_0(\text{HCl})\) provides the hydrogen ion (H⁺) and chloride ion (Cl⁻).
- \(\Lambda_0(\text{NaCl})\) provides sodium ion (Na⁺) and chloride ion (Cl⁻).

5. **Simplify the equation**:
By substituting the values, we can see that the Na⁺ and Cl⁻ ions cancel out, leaving us with:
\[ \Lambda_0(\text{CH}_3\text{COOH}) = \Lambda_0(\text{CH}_3\text{COO}^-) + \Lambda_0(\text{H}^+) \]

6. **Select the correct option**:
From the options provided, the one that corresponds to the equation derived above is:
\[ \text{Option 2: } \Lambda_0(\text{CH}_3\text{COONa}) + \Lambda_0(\text{HCl}) - \Lambda_0(\text{NaCl}) \]
Thus, the correct answer is option number 2.

### Final Answer:
The limiting molar conductivity of acetic acid (CH₃COOH) is given by option number 2.

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