At `25^(@)C`, the molar conductances at infinite dilution for the strong electrolytes `NaOH, NaCl " and " BaCl_(2) " are " 248 xx 10^(-4), 126 xx 10^(-4)" and " 280 xx 10^(-4) S m^(2) mol^(-1)` respectively. `Lamda_(m)^(@) Ba (OH)_(2) " in " S m^(2) mol^(-1)`

Options

A. `52.4 xx 10^(-4)`
B. `524 xx 10^(-4)`
C. `402 xx 10^(-4)`
D. `262 xx 10^(-4)`

Explanation

To find the molar conductance at infinite dilution (\( \Lambda_m^{\infty} \)) of barium hydroxide (\( \text{Ba(OH)}_2 \)), we can use the following relationship based on the molar conductances of the constituent ions:

\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = \Lambda_m^{\infty}(\text{BaCl}_2) + 2 \Lambda_m^{\infty}(\text{NaOH}) - \Lambda_m^{\infty}(\text{NaCl}) \]

### Step-by-step solution:

1. **Identify the given values:**
- \( \Lambda_m^{\infty}(\text{NaOH}) = 248 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \)
- \( \Lambda_m^{\infty}(\text{NaCl}) = 126 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \)
- \( \Lambda_m^{\infty}(\text{BaCl}_2) = 280 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \)

2. **Substitute the values into the equation:**
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = 280 \times 10^{-4} + 2 \times 248 \times 10^{-4} - 126 \times 10^{-4} \]

3. **Calculate \( 2 \times \Lambda_m^{\infty}(\text{NaOH}) \):**
\[ 2 \times 248 \times 10^{-4} = 496 \times 10^{-4} \]

4. **Now substitute this back into the equation:**
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = 280 \times 10^{-4} + 496 \times 10^{-4} - 126 \times 10^{-4} \]

5. **Combine the terms:**
\[ = (280 + 496 - 126) \times 10^{-4} \]
\[ = 650 \times 10^{-4} \]

6. **Final calculation:**
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = 650 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \]

### Final Answer:
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = 650 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \]

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