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 \( \text{Ba(OH)}_2 \), we can use the principle of Kohlrausch's law, which states that the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar conductivities of its individual ions.

### Step-by-step Solution:

1. **Identify the ions in \( \text{Ba(OH)}_2 \)**:
- The dissociation of \( \text{Ba(OH)}_2 \) in water can be represented as:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]
- Therefore, the ions are \( \text{Ba}^{2+} \) and \( 2 \text{OH}^- \).

2. **Write the expression for molar conductivity**:
- According to Kohlrausch's law:
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = \Lambda_m^{\infty}(\text{Ba}^{2+}) + 2 \Lambda_m^{\infty}(\text{OH}^-) \]

3. **Use the provided data**:
- We have the following molar conductivities at infinite dilution:
- \( \Lambda_m^{\infty}(\text{BaCl}_2) = 280 \times 10^{-4} \, \text{S m}^2/\text{mol} \)
- \( \Lambda_m^{\infty}(\text{NaCl}) = 126.5 \times 10^{-4} \, \text{S m}^2/\text{mol} \)
- \( \Lambda_m^{\infty}(\text{NaOH}) = 48 \times 10^{-4} \, \text{S m}^2/\text{mol} \)

4. **Extract the required ion conductivities**:
- For \( \text{BaCl}_2 \):
- The ions are \( \text{Ba}^{2+} \) and \( 2 \text{Cl}^- \):
\[ \Lambda_m^{\infty}(\text{Ba}^{2+}) + 2 \Lambda_m^{\infty}(\text{Cl}^-) = 280 \times 10^{-4} \]
- For \( \text{NaCl} \):
- The ions are \( \text{Na}^+ \) and \( \text{Cl}^- \):
\[ \Lambda_m^{\infty}(\text{Na}^+) + \Lambda_m^{\infty}(\text{Cl}^-) = 126.5 \times 10^{-4} \]
- For \( \text{NaOH} \):
- The ions are \( \text{Na}^+ \) and \( \text{OH}^- \):
\[ \Lambda_m^{\infty}(\text{Na}^+) + \Lambda_m^{\infty}(\text{OH}^-) = 48 \times 10^{-4} \]

5. **Solve the equations**:
- From the \( \text{NaCl} \) equation, we can express \( \Lambda_m^{\infty}(\text{Cl}^-) \):
\[ \Lambda_m^{\infty}(\text{Cl}^-) = 126.5 \times 10^{-4} - \Lambda_m^{\infty}(\text{Na}^+) \]
- Substitute this into the \( \text{BaCl}_2 \) equation:
\[ \Lambda_m^{\infty}(\text{Ba}^{2+}) + 2(126.5 \times 10^{-4} - \Lambda_m^{\infty}(\text{Na}^+)) = 280 \times 10^{-4} \]
- Rearranging gives:
\[ \Lambda_m^{\infty}(\text{Ba}^{2+}) - 2\Lambda_m^{\infty}(\text{Na}^+) = 280 \times 10^{-4} - 253 \times 10^{-4} \]
\[ \Lambda_m^{\infty}(\text{Ba}^{2+}) - 2\Lambda_m^{\infty}(\text{Na}^+) = 27 \times 10^{-4} \]

6. **Substituting \( \Lambda_m^{\infty}(\text{Na}^+) \)**:
- From the \( \text{NaOH} \) equation:
\[ \Lambda_m^{\infty}(\text{Na}^+) = 48 \times 10^{-4} - \Lambda_m^{\infty}(\text{OH}^-) \]
- Substitute this into the previous equation and solve for \( \Lambda_m^{\infty}(\text{Ba}^{2+}) \) and \( \Lambda_m^{\infty}(\text{OH}^-) \).

7. **Final calculation**:
- After calculating the values, we find:
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = 280 \times 10^{-4} + 2(48 \times 10^{-4}) - 2(126.5 \times 10^{-4}) \]
- Simplifying gives:
\[ \Lambda_m^{\infty}(\text{Ba(OH)}_2) = 280 \times 10^{-4} + 96 \times 10^{-4} - 253 \times 10^{-4} = 523 \times 10^{-4} \, \text{S m}^2/\text{mol} \]

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