From the following molar conductivities at infinite dilution,
`Lamda_(m)^(@) " for " Al_(2) (SO_(4))_(3) = 858 S cm^(2) mol^(-1)`
`Lamda_(m)^(@) " for " NH_(4)OH = 238.3 S cm^(2) mol^(-1)`
`Lamda_(m)^(@) " for " (NH_(4))_(2)SO_(4) = 238.4 S cm^(2) mol^(-1)`
Calculate `Lamda_(m)^(@) " for " Al(OH)_(3)`

Options

A. `715.2 S cm^(2) mol^(-1)`
B. `1575.6 S cm^(2) mol^(-1)`
C. `786.3 S cm^(2) mol^(-1)`
D. `157.56 S cm^(2) mol^(-1)`

Answer

To calculate the limiting molar conductivity (\( \Lambda_m^\circ \)) for \( \text{Al(OH)}_3 \) using the provided molar conductivities at infinite dilution, we can apply Kohlrausch's law, which states that the molar conductivity of an electrolyte can be expressed as the sum of the contributions from its individual ions.

### Step-by-Step Solution:

1. **Identify the ions in \( \text{Al(OH)}_3 \)**:
- The dissociation of \( \text{Al(OH)}_3 \) can be represented as:
\[ \text{Al(OH)}_3 \rightarrow \text{Al}^{3+} + 3 \text{OH}^- \]
- Therefore, the ions involved are \( \text{Al}^{3+} \) and \( \text{OH}^- \).

2. **Use the molar conductivities of the relevant ions**:
- From the given data:
- \( \Lambda_m^\circ \) for \( \text{Al}_2(\text{SO}_4)_3 \) = 858 S cm² mol⁻¹
- \( \Lambda_m^\circ \) for \( \text{NH}_4\text{OH} \) = 238.3 S cm² mol⁻¹
- \( \Lambda_m^\circ \) for \( (NH_4)_2SO_4 \) = 238.4 S cm² mol⁻¹

3. **Determine the contribution of \( \text{Al}^{3+} \)**:
- The molar conductivity of \( \text{Al}^{3+} \) can be derived from \( \text{Al}_2(\text{SO}_4)_3 \):
\[ \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \]
- Therefore, the contribution of \( \text{Al}^{3+} \) is:
\[ \Lambda_{\text{Al}^{3+}} = \frac{1}{2} \Lambda_m^\circ(\text{Al}_2(\text{SO}_4)_3) = \frac{858}{2} = 429 \text{ S cm}^2 \text{ mol}^{-1} \]

4. **Determine the contribution of \( \text{OH}^- \)**:
- The \( \text{OH}^- \) ions come from \( \text{NH}_4\text{OH} \):
- Since \( \text{NH}_4\text{OH} \) produces \( \text{NH}_4^+ \) and \( \text{OH}^- \):
\[ \text{NH}_4\text{OH} \rightarrow \text{NH}_4^+ + \text{OH}^- \]
- Therefore, the contribution of \( \text{OH}^- \) is:
\[ \Lambda_{\text{OH}^-} = \Lambda_m^\circ(\text{NH}_4\text{OH}) = 238.3 \text{ S cm}^2 \text{ mol}^{-1} \]

5. **Calculate the total molar conductivity for \( \text{Al(OH)}_3 \)**:
- The total molar conductivity for \( \text{Al(OH)}_3 \) is given by:
\[ \Lambda_m^\circ(\text{Al(OH)}_3) = \Lambda_{\text{Al}^{3+}} + 3 \Lambda_{\text{OH}^-} \]
- Substituting the values:
\[ \Lambda_m^\circ(\text{Al(OH)}_3) = 429 + 3 \times 238.3 \]
\[ = 429 + 714.9 = 1143.9 \text{ S cm}^2 \text{ mol}^{-1} \]

6. **Final Result**:
- The limiting molar conductivity for \( \text{Al(OH)}_3 \) is:
\[ \Lambda_m^\circ(\text{Al(OH)}_3) = 1143.9 \text{ S cm}^2 \text{ mol}^{-1} \]