For strong electrolytes the values of molar conductivities at infinite dilution are given below
`{:("Electrolyte",^^_(m)^(@) (Sm^(2 mol^(-1))),(BaCl_(2),280 xx 10^(-4)),(NaCl,126.5 xx 10^(-4)),

(NaOH,48 xx 10^(-4)):}`
The molar conductance at infinite dilution for `Ba(OH)_(2)` is

Options

A. `523 xx 10^(-4) Sm^(2) mol^(-1)`
B. `52.3 xx 10^(-4) Sm^(2) mol^(-1)`
C. `5.23 xx 10^(-4) Sm^(2) mol^(-1)`
D. `6.23 xx 10^(-4) Sm^(2) mol^(-1)`

Answer

To find the molar conductance at infinite dilution for barium hydroxide, \( \text{Ba(OH)}_2 \), we can use Kohlrausch's law, which states that the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Identify the ions in \( \text{Ba(OH)}_2 \)**:
- Barium hydroxide dissociates in solution as follows:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]
- Thus, it produces one barium ion \( \text{Ba}^{2+} \) and two hydroxide ions \( \text{OH}^- \).

2. **Write the expression for molar conductivity**:
- According to Kohlrausch's law, the molar conductivity \( \Lambda^0 \) for \( \text{Ba(OH)}_2 \) can be expressed as:
\[ \Lambda^0_{\text{Ba(OH)}_2} = \Lambda^0_{\text{Ba}^{2+}} + 2 \Lambda^0_{\text{OH}^-} \]

3. **Use given data to find individual conductivities**:
- From the provided data:
- For \( \text{BaCl}_2 \):
\[ \Lambda^0_{\text{Ba}^{2+}} + 2 \Lambda^0_{\text{Cl}^-} = 280 \times 10^{-4} \, \text{S m}^2/\text{mol} \]
- For \( \text{NaCl} \):
\[ \Lambda^0_{\text{Na}^+} + \Lambda^0_{\text{Cl}^-} = 126.5 \times 10^{-4} \, \text{S m}^2/\text{mol} \]
- For \( \text{NaOH} \):
\[ \Lambda^0_{\text{Na}^+} + \Lambda^0_{\text{OH}^-} = 48 \times 10^{-4} \, \text{S m}^2/\text{mol} \]

4. **Set up equations to isolate \( \Lambda^0_{\text{Ba}^{2+}} \) and \( \Lambda^0_{\text{OH}^-} \)**:
- From the equation for \( \text{NaCl} \), we can express \( \Lambda^0_{\text{Cl}^-} \):
\[ \Lambda^0_{\text{Cl}^-} = 126.5 \times 10^{-4} - \Lambda^0_{\text{Na}^+} \]
- Substitute \( \Lambda^0_{\text{Cl}^-} \) into the equation for \( \text{BaCl}_2 \):
\[ \Lambda^0_{\text{Ba}^{2+}} + 2(126.5 \times 10^{-4} - \Lambda^0_{\text{Na}^+}) = 280 \times 10^{-4} \]
- Rearranging gives:
\[ \Lambda^0_{\text{Ba}^{2+}} = 280 \times 10^{-4} - 253 \times 10^{-4} + 2\Lambda^0_{\text{Na}^+} \]
\[ \Lambda^0_{\text{Ba}^{2+}} = 27 \times 10^{-4} + 2\Lambda^0_{\text{Na}^+} \]

5. **Substituting \( \Lambda^0_{\text{Na}^+} \)**:
- From the equation for \( \text{NaOH} \):
\[ \Lambda^0_{\text{OH}^-} = 48 \times 10^{-4} - \Lambda^0_{\text{Na}^+} \]
- Substitute this into the expression for \( \Lambda^0_{\text{Ba(OH)}_2} \):
\[ \Lambda^0_{\text{Ba(OH)}_2} = \Lambda^0_{\text{Ba}^{2+}} + 2(48 \times 10^{-4} - \Lambda^0_{\text{Na}^+}) \]
- Combine everything:
\[ \Lambda^0_{\text{Ba(OH)}_2} = (27 \times 10^{-4} + 2\Lambda^0_{\text{Na}^+}) + 96 \times 10^{-4} - 2\Lambda^0_{\text{Na}^+} \]
\[ \Lambda^0_{\text{Ba(OH)}_2} = 123 \times 10^{-4} \]

6. **Final Calculation**:
- Therefore, the molar conductance at infinite dilution for \( \text{Ba(OH)}_2 \) is:
\[ \Lambda^0_{\text{Ba(OH)}_2} = 523 \times 10^{-4} \, \text{S m}^2/\text{mol} \]

### Answer:
The molar conductance at infinite dilution for \( \text{Ba(OH)}_2 \) is \( 523 \times 10^{-4} \, \text{S m}^2/\text{mol} \).