According of Kohlrausch law, the limiting value of molar conductivity of an electrolyte `A_(2)B` is
(a)`lambda^(oo) ._(A^(+)) + lambda^(oo) ._((B^(-))`
(b)`lambda^(oo) ._(A^(+)) - lambda^(oo) ._(B^(-))`
(c)`2lambda^(oo) ._(A^(+)) +(1)/(2)lambda^(oo) ._((B^(-))`
(d)`2lambda^(oo) ._(A^(+)) + lambda^(oo) ._(B^(-))`

Options

A. `lambda^(oo) ._(A^(+)) + lambda^(oo) ._((B^(-))`
B. `lambda^(oo) ._(A^(+)) - lambda^(oo) ._(B^(-))`
C. `2lambda^(oo) ._(A^(+)) +(1)/(2)lambda^(oo) ._((B^(-))`
D. `2lambda^(oo) ._(A^(+)) + lambda^(oo) ._(B^(-))`

Explanation

To solve the question regarding the limiting value of molar conductivity of the electrolyte \(A_2B\) according to Kohlrausch's law, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electrolyte Dissociation:**
The electrolyte \(A_2B\) dissociates in solution to form ions. The dissociation can be represented as:
\[ A_2B \rightarrow 2A^+ + B^{2-} \]
This means that for each formula unit of \(A_2B\), we get 2 moles of \(A^+\) ions and 1 mole of \(B^{2-}\) ions.

2. **Applying Kohlrausch's Law:**
According to Kohlrausch's law, the molar conductivity \(\Lambda^{\infty}\) of an electrolyte at infinite dilution is the sum of the contributions from its cations and anions:
\[ \Lambda^{\infty}(A_2B) = \nu_+ \lambda^{\infty}(A^+) + \nu_- \lambda^{\infty}(B^{2-}) \]
where \(\nu_+\) and \(\nu_-\) are the stoichiometric coefficients of the cation and anion, respectively, and \(\lambda^{\infty}\) represents the molar conductivity at infinite dilution.

3. **Identifying Stoichiometric Coefficients:**
From the dissociation reaction, we see that:
- The stoichiometric coefficient for \(A^+\) is \(2\) (since there are 2 moles of \(A^+\)).
- The stoichiometric coefficient for \(B^{2-}\) is \(1\) (since there is 1 mole of \(B^{2-}\)).

4. **Writing the Expression:**
Substituting the stoichiometric coefficients into the equation, we get:
\[ \Lambda^{\infty}(A_2B) = 2\lambda^{\infty}(A^+) + 1\lambda^{\infty}(B^{2-}) \]
This simplifies to:
\[ \Lambda^{\infty}(A_2B) = 2\lambda^{\infty}(A^+) + \lambda^{\infty}(B^{2-}) \]

5. **Choosing the Correct Option:**
Now, we look at the options provided:
- (a) \(\lambda^{\infty}(A^+) + \lambda^{\infty}(B^{-})\)
- (b) \(\lambda^{\infty}(A^+) - \lambda^{\infty}(B^{-})\)
- (c) \(2\lambda^{\infty}(A^+) + \frac{1}{2}\lambda^{\infty}(B^{-})\)
- (d) \(2\lambda^{\infty}(A^+) + \lambda^{\infty}(B^{-})\)

The expression we derived matches option (d).

### Final Answer:
The correct answer is (d) \(2\lambda^{\infty}(A^+) + \lambda^{\infty}(B^{2-})\).