At `25^(@)`C, the molar conductance 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 m2 mol^(-1)` respectively. The value of `wedge_(m)^(infty) Ba(OH)_(2)` in S `m^(2) mol^(-1)` will be:

Options

A. `52.xx10^(-4)`
B. `524xx10^(-4)`
C. `402xx10^(-4)`
D. `262xx10^(-4)`

Answer

To find the molar conductance at infinite dilution for barium hydroxide, \( \Lambda_m^{\infty} \) for \( Ba(OH)_2 \), we can use the molar conductance values of the given strong electrolytes \( NaOH \), \( NaCl \), and \( BaCl_2 \).

### Step-by-Step Solution:

1. **Identify the dissociation of the compounds**:
- \( NaOH \) dissociates into \( Na^+ \) and \( OH^- \).
- \( NaCl \) dissociates into \( Na^+ \) and \( Cl^- \).
- \( BaCl_2 \) dissociates into \( Ba^{2+} \) and \( 2Cl^- \).
- \( Ba(OH)_2 \) dissociates into \( Ba^{2+} \) and \( 2OH^- \).

2. **Set up the equation**:
To find the molar conductance of \( Ba(OH)_2 \), we can use the following relationship:
\[ \Lambda_m^{\infty} (Ba(OH)_2) = 2\Lambda_m^{\infty} (NaOH) + \Lambda_m^{\infty} (BaCl_2) - 2\Lambda_m^{\infty} (NaCl) \]

3. **Substitute the given values**:
- \( \Lambda_m^{\infty} (NaOH) = 248 \times 10^{-4} \, S \, m^2 \, mol^{-1} \)
- \( \Lambda_m^{\infty} (NaCl) = 126 \times 10^{-4} \, S \, m^2 \, mol^{-1} \)
- \( \Lambda_m^{\infty} (BaCl_2) = 280 \times 10^{-4} \, S \, m^2 \, mol^{-1} \)

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

4. **Calculate each term**:
- \( 2(248 \times 10^{-4}) = 496 \times 10^{-4} \)
- \( 2(126 \times 10^{-4}) = 252 \times 10^{-4} \)

5. **Combine the values**:
\[ \Lambda_m^{\infty} (Ba(OH)_2) = 496 \times 10^{-4} + 280 \times 10^{-4} - 252 \times 10^{-4} \]
\[ = (496 + 280 - 252) \times 10^{-4} \]
\[ = 524 \times 10^{-4} \, S \, m^2 \, mol^{-1} \]

6. **Final result**:
\[ \Lambda_m^{\infty} (Ba(OH)_2) = 524 \times 10^{-4} \, S \, m^2 \, mol^{-1} \]

### Conclusion:
The value of \( \Lambda_m^{\infty} (Ba(OH)_2) \) is \( 524 \times 10^{-4} \, S \, m^2 \, mol^{-1} \).