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 conductivities of the strong electrolytes provided: NaOH, NaCl, and BaCl₂.

### Step-by-Step Solution:

1. **Identify the dissociation of Ba(OH)₂**:
Ba(OH)₂ dissociates into:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

2. **Use the known molar ionic conductivities**:
- Molar ionic conductivity of NaOH (λ₀m(NaOH)) = 250 mho cm² mol⁻¹
- Molar ionic conductivity of NaCl (λ₀m(NaCl)) = 125 mho cm² mol⁻¹
- Molar ionic conductivity of BaCl₂ (λ₀m(BaCl₂)) = 300 mho cm² mol⁻¹

3. **Set up the equation for molar conductivity of Ba(OH)₂**:
The molar conductivity (λ₀m) of Ba(OH)₂ can be expressed in terms of the molar conductivities of the other electrolytes:
\[ \lambda^0_m(\text{Ba(OH)}_2) = \lambda^0_m(\text{BaCl}_2) + 2 \lambda^0_m(\text{NaOH}) - 2 \lambda^0_m(\text{NaCl}) \]

4. **Substitute the values into the equation**:
\[ \lambda^0_m(\text{Ba(OH)}_2) = 300 + 2(250) - 2(125) \]

5. **Calculate the values**:
- First, calculate \(2(250) = 500\)
- Then, calculate \(2(125) = 250\)
- Now substitute these values back into the equation:
\[ \lambda^0_m(\text{Ba(OH)}_2) = 300 + 500 - 250 \]
\[ \lambda^0_m(\text{Ba(OH)}_2) = 300 + 250 = 550 \text{ mho cm}^2 \text{ mol}^{-1} \]

### Final Answer:
The molar conductivity of Ba(OH)₂ at infinite dilution is **550 mho cm² mol⁻¹**.

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