`wedge^(@)._(aq)` of `BaCl_(2), H_(2)SO_(4)`, and `HCl(aq)` solutions are `x_(1), x_(2),` and `x_(3),` respectively . `wedge^(@)._(m(BaSO_(4)))` is `:`

Options

A. `x_(1)+x_(2)-x_(3)`
B. `x_(1)-x_(2)-x_(3)`
C. `x_(1)+x_(2)-2x_(3)`
D. `x_(1)-2x_(2)+x_(3)`

Explanation

To find the molar conductivity at infinite dilution for BaSO4, we can use Kohlrausch's law, which states that the molar conductivity at infinite dilution of an electrolyte can be expressed in terms of the molar conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Identify the Molar Conductivities**:
- Let the molar conductivity at infinite dilution for BaCl2 be denoted as \( \lambda_{BaCl_2} = x_1 \).
- Let the molar conductivity at infinite dilution for H2SO4 be denoted as \( \lambda_{H_2SO_4} = x_2 \).
- Let the molar conductivity at infinite dilution for HCl be denoted as \( \lambda_{HCl} = x_3 \).

2. **Write the Molar Conductivity Expressions**:
- For BaCl2, which dissociates into Ba²⁺ and 2 Cl⁻:
\[ \lambda_{BaCl_2} = \lambda_{Ba^{2+}} + 2\lambda_{Cl^-} \]
Thus, we can express this as:
\[ x_1 = \lambda_{Ba^{2+}} + 2\lambda_{Cl^-} \quad \text{(Equation 1)} \]

- For H2SO4, which dissociates into 2 H⁺ and SO₄²⁻:
\[ \lambda_{H_2SO_4} = 2\lambda_{H^+} + \lambda_{SO_4^{2-}} \]
Thus, we can express this as:
\[ x_2 = 2\lambda_{H^+} + \lambda_{SO_4^{2-}} \quad \text{(Equation 2)} \]

- For HCl, which dissociates into H⁺ and Cl⁻:
\[ \lambda_{HCl} = \lambda_{H^+} + \lambda_{Cl^-} \]
Thus, we can express this as:
\[ x_3 = \lambda_{H^+} + \lambda_{Cl^-} \quad \text{(Equation 3)} \]

3. **Set Up the Equation for BaSO4**:
- BaSO4 dissociates into Ba²⁺ and SO₄²⁻:
\[ \lambda_{BaSO_4} = \lambda_{Ba^{2+}} + \lambda_{SO_4^{2-}} \quad \text{(Equation 4)} \]

4. **Combine the Equations**:
- We can add Equation 1 and Equation 2, and then subtract twice Equation 3:
\[ \lambda_{BaSO_4} = \lambda_{Ba^{2+}} + \lambda_{SO_4^{2-}} = (x_1 - 2\lambda_{Cl^-}) + (x_2 - 2\lambda_{H^+}) \quad \text{(from Equations 1, 2, and 3)} \]
- Rearranging gives us:
\[ \lambda_{BaSO_4} = x_1 + x_2 - 2x_3 \]

5. **Final Result**:
- Therefore, the molar conductivity at infinite dilution for BaSO4 is:
\[ \lambda_{BaSO_4} = x_1 + x_2 - 2x_3 \]

### Conclusion:
The answer is \( x_1 + x_2 - 2x_3 \).