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 molar conductivity at infinite dilution (Λ_m^(@)) for Al(OH)₃ using the provided molar conductivities of the other compounds, we can follow these steps:

### Step 1: Identify the ions in Al(OH)₃
Al(OH)₃ dissociates into:
- 1 Al³⁺ ion
- 3 OH⁻ ions

### Step 2: Write the expression for Λ_m^(@) of Al(OH)₃
Using the molar conductivities of the ions, we can express the molar conductivity of Al(OH)₃ as:
\[ Λ_m^(@) \text{(Al(OH)₃)} = Λ_m^(@) \text{(Al³⁺)} + 3 \cdot Λ_m^(@) \text{(OH⁻)} \]

### Step 3: Use the provided data to find Λ_m^(@) for Al³⁺ and OH⁻
From the problem, we know:
- For Al₂(SO₄)₃:
\[ Λ_m^(@) \text{(Al₂(SO₄)₃)} = 858 \, \text{S cm}^2 \text{mol}^{-1} \]
This dissociates into 2 Al³⁺ and 3 SO₄²⁻ ions:
\[ Λ_m^(@) \text{(Al³⁺)} + 3 \cdot Λ_m^(@) \text{(SO₄²⁻)} = 858 \]

- For (NH₄)₂SO₄:
\[ Λ_m^(@) \text{((NH₄)₂SO₄)} = 238.4 \, \text{S cm}^2 \text{mol}^{-1} \]
This dissociates into 2 NH₄⁺ and 1 SO₄²⁻:
\[ 2 \cdot Λ_m^(@) \text{(NH₄⁺)} + Λ_m^(@) \text{(SO₄²⁻)} = 238.4 \]

- For NH₄OH:
\[ Λ_m^(@) \text{(NH₄OH)} = 238.3 \, \text{S cm}^2 \text{mol}^{-1} \]
This dissociates into NH₄⁺ and OH⁻:
\[ Λ_m^(@) \text{(NH₄⁺)} + Λ_m^(@) \text{(OH⁻)} = 238.3 \]

### Step 4: Solve the equations
1. From (NH₄)₂SO₄:
\[ 2 \cdot Λ_m^(@) \text{(NH₄⁺)} + Λ_m^(@) \text{(SO₄²⁻)} = 238.4 \quad \text{(Equation 1)} \]

2. From Al₂(SO₄)₃:
\[ Λ_m^(@) \text{(Al³⁺)} + 3 \cdot Λ_m^(@) \text{(SO₄²⁻)} = 858 \quad \text{(Equation 2)} \]

3. From NH₄OH:
\[ Λ_m^(@) \text{(NH₄⁺)} + Λ_m^(@) \text{(OH⁻)} = 238.3 \quad \text{(Equation 3)} \]

### Step 5: Express Λ_m^(@) for Al(OH)₃
Using the values from the equations, we can express:
\[ Λ_m^(@) \text{(Al(OH)₃)} = \frac{1}{2} Λ_m^(@) \text{(Al₂(SO₄)₃)} + 3 Λ_m^(@) \text{(NH₄OH)} - \frac{3}{2} Λ_m^(@) \text{((NH₄)₂SO₄)} \]

### Step 6: Substitute the values
Substituting the known values:
\[ Λ_m^(@) \text{(Al(OH)₃)} = \frac{1}{2} \cdot 858 + 3 \cdot 238.3 - \frac{3}{2} \cdot 238.4 \]

Calculating:
\[ = 429 + 714.9 - 357.6 \]
\[ = 429 + 714.9 - 357.6 = 786.3 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
\[ Λ_m^(@) \text{(Al(OH)₃)} = 786.3 \, \text{S cm}^2 \text{mol}^{-1} \]