Equivalent conductance of `BaCI_(2),H_(2)SO_(4)` & `HCI` at infinite are `A_(oo)^(1),A_(oo)^(2)& A_(oo)^(3)` conductance of `BaSO_(4)` solution is:

Options

A. `A_(oo)^(1) +A_(oo)^(2)-2 A_(oo)^(3)`
B. `A_(oo)^(2) +A_(oo)^(2)-2 A_(oo)^(3)`
C. `A_(oo)^(1) +A_(oo)^(2) +A_(oo)^(3)`
D. `A_(oo)^(1) +A_(oo)^(2)-A_(oo)^(3)`

Explanation

To find the equivalent conductance of BaSO4 at infinite dilution, we can use the concept of equivalent conductance at infinite dilution and apply Kaldoroff's law. Here’s how we can solve the problem step by step:

### Step-by-Step Solution

1. **Identify the Equivalent Conductance Expressions**:
- For BaCl2:
\[ \Lambda_{BaCl_2}^\infty = \Lambda_{Ba^{2+}}^\infty + 2 \Lambda_{Cl^-}^\infty \]
This is given as \( A_1^\infty \).

- For H2SO4:
\[ \Lambda_{H_2SO_4}^\infty = 2 \Lambda_{H^+}^\infty + \Lambda_{SO_4^{2-}}^\infty \]
This is given as \( A_2^\infty \).

- For HCl:
\[ \Lambda_{HCl}^\infty = \Lambda_{H^+}^\infty + \Lambda_{Cl^-}^\infty \]
This is given as \( A_3^\infty \).

2. **Set Up the Equations**:
- From the above, we have:
\[ \Lambda_{Ba^{2+}}^\infty + 2 \Lambda_{Cl^-}^\infty = A_1^\infty \quad \text{(1)} \]
\[ 2 \Lambda_{H^+}^\infty + \Lambda_{SO_4^{2-}}^\infty = A_2^\infty \quad \text{(2)} \]
\[ \Lambda_{H^+}^\infty + \Lambda_{Cl^-}^\infty = A_3^\infty \quad \text{(3)} \]

3. **Manipulate the Equations**:
- We need to find \( \Lambda_{BaSO_4}^\infty \):
\[ \Lambda_{BaSO_4}^\infty = \Lambda_{Ba^{2+}}^\infty + \Lambda_{SO_4^{2-}}^\infty \]
- To eliminate \( \Lambda_{Cl^-}^\infty \) and \( \Lambda_{H^+}^\infty \), we can manipulate equations (1), (2), and (3).

4. **Express \( \Lambda_{Cl^-}^\infty \) and \( \Lambda_{H^+}^\infty \)**:
- From equation (3):
\[ \Lambda_{H^+}^\infty = A_3^\infty - \Lambda_{Cl^-}^\infty \quad \text{(4)} \]
- Substitute (4) into equation (2):
\[ 2(A_3^\infty - \Lambda_{Cl^-}^\infty) + \Lambda_{SO_4^{2-}}^\infty = A_2^\infty \]
Rearranging gives:
\[ \Lambda_{SO_4^{2-}}^\infty = A_2^\infty - 2A_3^\infty + 2\Lambda_{Cl^-}^\infty \quad \text{(5)} \]

5. **Substituting Back**:
- Substitute (5) into (1):
\[ \Lambda_{Ba^{2+}}^\infty + 2\Lambda_{Cl^-}^\infty = A_1^\infty \]
- Now we have two equations (1) and (5) to work with.

6. **Final Calculation**:
- We can now express \( \Lambda_{BaSO_4}^\infty \):
\[ \Lambda_{BaSO_4}^\infty = \Lambda_{Ba^{2+}}^\infty + \Lambda_{SO_4^{2-}}^\infty \]
- By substituting the values from equations (1) and (5), we can derive:
\[ \Lambda_{BaSO_4}^\infty = A_1^\infty + A_2^\infty - 2A_3^\infty \]

### Final Answer
Thus, the equivalent conductance of BaSO4 at infinite dilution is:
\[ \Lambda_{BaSO_4}^\infty = A_1^\infty + A_2^\infty - 2A_3^\infty \]