The quivalnet conductance of any electrolyte `MA` at infinite dilution `wedge^(@)._((MA))` is equal `(` more than one correct answer`)`
1) `wedge^(@)._((MA))=wedge^(c-)._((MCl))+wedge^(@)._((NaA))+wedge^(@)._((NaCl))`
2) `wedge^(@)._((MA))=wedge^(c-)._((MCl))+wedge^(@)._((NaA))-wedge^(@)._((NaCl))`
3) `wedge^(@)._((MA))=lambda^(@)._((M^(o+)))+lambda^(@)._((A^(c-)))`
4) `wedge^(@)._((MA))=wedge^(c-)._((MCl))+wedge^(@)._((NaA))-wedge^(@)._((NaCl))`

Options

A. `wedge^(@)._((MA))=wedge^(c-)._((MCl))+wedge^(@)._((NaA))+wedge^(@)._((NaCl))`
B. `wedge^(@)._((MA))=wedge^(c-)._((MCl))+wedge^(@)._((NaA))-wedge^(@)._((NaCl))`
C. `wedge^(@)._((MA))=lambda^(@)._((M^(o+)))+lambda^(@)._((A^(c-)))`
D. `wedge^(@)._((MA))=wedge^(c-)._((MCl))+wedge^(@)._((NaA))-wedge^(@)._((NaCl))`

Explanation

To solve the question regarding the equivalent conductance of the electrolyte \( MA \) at infinite dilution, we will analyze the options provided based on Kohlrausch's Law.

### Step-by-Step Solution:

1. **Understanding Equivalent Conductance**:
The equivalent conductance \( \Lambda^0_{MA} \) at infinite dilution is defined as the conductance of one equivalent of the electrolyte when it is completely dissociated into its ions.

2. **Kohlrausch's Law**:
According to Kohlrausch's Law, the molar conductivity of a weak electrolyte at infinite dilution can be expressed as the sum of the molar conductivities of its constituent ions:
\[ \Lambda^0_{MA} = \Lambda^0_{M^+} + \Lambda^0_{A^-} \]
where \( M^+ \) is the cation and \( A^- \) is the anion.

3. **Evaluating the Options**:
- **Option 1**:
\[ \Lambda^0_{MA} = \Lambda^0_{MCl} + \Lambda^0_{NaA} + \Lambda^0_{NaCl} \]
This does not match our expression from Kohlrausch's Law, as it includes additional terms that do not simplify to \( \Lambda^0_{M^+} + \Lambda^0_{A^-} \). **Not correct**.

- **Option 2**:
\[ \Lambda^0_{MA} = \Lambda^0_{MCl} + \Lambda^0_{NaA} - \Lambda^0_{NaCl} \]
Breaking this down:
- \( \Lambda^0_{MCl} = \Lambda^0_{M^+} + \Lambda^0_{Cl^-} \)
- \( \Lambda^0_{NaA} = \Lambda^0_{Na^+} + \Lambda^0_{A^-} \)
- \( \Lambda^0_{NaCl} = \Lambda^0_{Na^+} + \Lambda^0_{Cl^-} \)

Substituting these into the equation, we find that the \( \Lambda^0_{Na^+} \) and \( \Lambda^0_{Cl^-} \) terms cancel out, leading to:
\[ \Lambda^0_{MA} = \Lambda^0_{M^+} + \Lambda^0_{A^-} \]
This matches our expression from Kohlrausch's Law. **Correct**.

- **Option 3**:
\[ \Lambda^0_{MA} = \Lambda^0_{M^+} + \Lambda^0_{A^-} \]
This directly follows from Kohlrausch's Law, confirming that it is indeed correct. **Correct**.

- **Option 4**:
\[ \Lambda^0_{MA} = \Lambda^0_{MCl} + \Lambda^0_{NaA} - \Lambda^0_{NaCl} \]
This is identical to Option 2, which we have already confirmed as correct. **Correct**.

4. **Conclusion**:
The correct options are 2, 3, and 4.

### Final Answer:
The equivalent conductance of the electrolyte \( MA \) at infinite dilution is correctly represented by options 2, 3, and 4.