At `25^@C` the molar conductance at infinite dilution for the strong electrolytes NaOH ,NaCl and` BaCl_2`are `280xx10^-4,126xx10^-4` and `280xx10^-4Sm^2mol^-1`respectively `wedge_m^0Ba(OH)_2` in `Sm^2mol^-1` will be

Options

A. `52.4xx10^-41`
B. `524xx10^-4`
C. `40.4xx10^-4`
D. `262xx10^-4`

Answer

To find the molar conductance at infinite dilution (\( \Lambda_m^0 \)) for barium hydroxide (\( Ba(OH)_2 \)), we can use the molar conductance values of the given strong electrolytes: \( NaOH \), \( NaCl \), and \( BaCl_2 \).

### Step-by-Step Solution:

1. **Identify the ions in \( Ba(OH)_2 \)**:
- \( Ba(OH)_2 \) dissociates into \( Ba^{2+} \) and \( 2OH^{-} \) ions.
- Therefore, \( \Lambda_m^0 \) for \( Ba(OH)_2 \) can be expressed as:
\[ \Lambda_m^0 (Ba(OH)_2) = \Lambda_m^0 (Ba^{2+}) + 2 \Lambda_m^0 (OH^{-}) \]

2. **Use the values of \( BaCl_2 \)**:
- The dissociation of \( BaCl_2 \) gives \( Ba^{2+} \) and \( 2Cl^{-} \).
- Thus, the molar conductance of \( BaCl_2 \) can be used to find \( \Lambda_m^0 (Ba^{2+}) \):
\[ \Lambda_m^0 (Ba^{2+}) = \Lambda_m^0 (BaCl_2) - 2 \Lambda_m^0 (Cl^{-}) \]

3. **Use the values of \( NaCl \)**:
- The molar conductance of \( NaCl \) gives \( Na^{+} \) and \( Cl^{-} \).
- Therefore, we can express \( \Lambda_m^0 (Cl^{-}) \) as:
\[ \Lambda_m^0 (Cl^{-}) = \Lambda_m^0 (NaCl) \]

4. **Substituting the values**:
- Given:
- \( \Lambda_m^0 (NaCl) = 126 \times 10^{-4} \, S m^2 mol^{-1} \)
- \( \Lambda_m^0 (BaCl_2) = 280 \times 10^{-4} \, S m^2 mol^{-1} \)
- \( \Lambda_m^0 (NaOH) = 280 \times 10^{-4} \, S m^2 mol^{-1} \)
- Substitute into the equation:
\[ \Lambda_m^0 (Ba^{2+}) = 280 \times 10^{-4} - 2 \times 126 \times 10^{-4} \]
\[ \Lambda_m^0 (Ba^{2+}) = 280 \times 10^{-4} - 252 \times 10^{-4} = 28 \times 10^{-4} \, S m^2 mol^{-1} \]

5. **Calculate \( \Lambda_m^0 (OH^{-}) \)**:
- From \( NaOH \):
\[ \Lambda_m^0 (OH^{-}) = \Lambda_m^0 (NaOH) = 280 \times 10^{-4} \, S m^2 mol^{-1} \]

6. **Final calculation for \( \Lambda_m^0 (Ba(OH)_2) \)**:
- Substitute back into the equation:
\[ \Lambda_m^0 (Ba(OH)_2) = 28 \times 10^{-4} + 2 \times 280 \times 10^{-4} \]
\[ \Lambda_m^0 (Ba(OH)_2) = 28 \times 10^{-4} + 560 \times 10^{-4} = 588 \times 10^{-4} \, S m^2 mol^{-1} \]

### Final Answer:
\[ \Lambda_m^0 (Ba(OH)_2) = 588 \times 10^{-4} \, S m^2 mol^{-1} \]