For the strong electrolytes `NaOH, NaCl and BaCl_(2)` the molar ionic conductivities at infinite dilution are 250, 125 and 300 `"mho cm"^(2)"mol"^(-1)` respectively. The molar conductivity of `Ba(OH)_(2)` at infinite dilution `("mho cm"^(2)"mol"^(-1))` is .

Answer

To find the molar conductivity of Ba(OH)₂ at infinite dilution, we can use the molar ionic conductivities of the strong electrolytes provided: NaOH, NaCl, and BaCl₂. The molar conductivity of Ba(OH)₂ can be calculated using the contributions from its constituent ions.

### Step-by-Step Solution:

1. **Identify the ions in Ba(OH)₂**:
- Ba(OH)₂ dissociates into Ba²⁺ and 2 OH⁻ ions.
- Thus, the total molar conductivity (Λ°) of Ba(OH)₂ can be expressed as:
\[ \Lambda°_{Ba(OH)₂} = \Lambda°_{Ba^{2+}} + 2 \Lambda°_{OH^{-}} \]

2. **Find the molar conductivity of Ba²⁺**:
- Ba²⁺ comes from BaCl₂. The dissociation of BaCl₂ gives one Ba²⁺ and two Cl⁻ ions.
- Therefore, we can express the molar conductivity of BaCl₂ as:
\[ \Lambda°_{BaCl₂} = \Lambda°_{Ba^{2+}} + 2 \Lambda°_{Cl^{-}} \]
- Given that the molar conductivity of BaCl₂ is 300 mho cm² mol⁻¹, we can write:
\[ 300 = \Lambda°_{Ba^{2+}} + 2 \Lambda°_{Cl^{-}} \quad \text{(1)} \]

3. **Find the molar conductivity of NaOH**:
- NaOH dissociates into Na⁺ and OH⁻ ions.
- Therefore, the molar conductivity of NaOH can be expressed as:
\[ \Lambda°_{NaOH} = \Lambda°_{Na^{+}} + \Lambda°_{OH^{-}} \]
- Given that the molar conductivity of NaOH is 250 mho cm² mol⁻¹, we can write:
\[ 250 = \Lambda°_{Na^{+}} + \Lambda°_{OH^{-}} \quad \text{(2)} \]

4. **Find the molar conductivity of NaCl**:
- NaCl dissociates into Na⁺ and Cl⁻ ions.
- Therefore, the molar conductivity of NaCl can be expressed as:
\[ \Lambda°_{NaCl} = \Lambda°_{Na^{+}} + \Lambda°_{Cl^{-}} \]
- Given that the molar conductivity of NaCl is 125 mho cm² mol⁻¹, we can write:
\[ 125 = \Lambda°_{Na^{+}} + \Lambda°_{Cl^{-}} \quad \text{(3)} \]

5. **Solve equations (2) and (3)**:
- From equation (3), we can express \(\Lambda°_{Na^{+}}\) as:
\[ \Lambda°_{Na^{+}} = 125 - \Lambda°_{Cl^{-}} \quad \text{(4)} \]
- Substitute equation (4) into equation (2):
\[ 250 = (125 - \Lambda°_{Cl^{-}}) + \Lambda°_{OH^{-}} \]
\[ 250 = 125 - \Lambda°_{Cl^{-}} + \Lambda°_{OH^{-}} \]
\[ \Lambda°_{OH^{-}} - \Lambda°_{Cl^{-}} = 125 \quad \text{(5)} \]

6. **Substitute equation (5) into equation (1)**:
- Substitute \(\Lambda°_{OH^{-}} = \Lambda°_{Cl^{-}} + 125\) into equation (1):
\[ 300 = \Lambda°_{Ba^{2+}} + 2(\Lambda°_{Cl^{-}}) \]
\[ 300 = \Lambda°_{Ba^{2+}} + 2\Lambda°_{Cl^{-}} \]
- Now substitute \(\Lambda°_{OH^{-}} = \Lambda°_{Cl^{-}} + 125\) into the equation for Ba(OH)₂:
\[ \Lambda°_{Ba(OH)₂} = \Lambda°_{Ba^{2+}} + 2(\Lambda°_{Cl^{-}} + 125) \]

7. **Calculate the final value**:
- From equation (1), we can express \(\Lambda°_{Ba^{2+}}\):
\[ \Lambda°_{Ba^{2+}} = 300 - 2\Lambda°_{Cl^{-}} \]
- Substitute this back into the equation for Ba(OH)₂:
\[ \Lambda°_{Ba(OH)₂} = (300 - 2\Lambda°_{Cl^{-}}) + 2(\Lambda°_{Cl^{-}} + 125) \]
\[ = 300 - 2\Lambda°_{Cl^{-}} + 2\Lambda°_{Cl^{-}} + 250 \]
\[ = 300 + 250 = 550 \text{ mho cm}² \text{ mol}^{-1} \]

Thus, the molar conductivity of Ba(OH)₂ at infinite dilution is **550 mho cm² mol⁻¹**.