Given
`lamda^(oo) [1//2 Mg^(2+)] = 53.06 ohm^(-1) cm^(2) mol^(-1)`
`lamda^(oo) [1//2 SO_(4)^(2-)] = 80 ohm^(-1) cm^(2) mol^(-1)`
`lamda^(oo) [1//3 Al^(3+)] = 63 ohm^(-1) cm^(2) mol^(-1)`
Calculate the values of `Lamda_(m)^(oo) [Al_(2) (SO_(4))_(3)] and Lamda^(oo) [Mg SO_(4)]`

Answer

To calculate the molar conductivity at infinite dilution (Λ_m^∞) for the compounds Al₂(SO₄)₃ and MgSO₄, we will use the given values of molar conductivities for the ions involved.

### Step-by-Step Solution:

1. **Identify the ions in the compounds:**
- For Al₂(SO₄)₃: The ions are 2 Al³⁺ and 3 SO₄²⁻.
- For MgSO₄: The ions are 1 Mg²⁺ and 1 SO₄²⁻.

2. **Write the formula for molar conductivity at infinite dilution:**
- For Al₂(SO₄)₃:
\[ Λ_m^∞ [Al₂(SO₄)₃] = 2 \cdot Λ^∞ [Al^{3+}] + 3 \cdot Λ^∞ [SO₄^{2-}] \]
- For MgSO₄:
\[ Λ_m^∞ [MgSO₄] = 1 \cdot Λ^∞ [Mg^{2+}] + 1 \cdot Λ^∞ [SO₄^{2-}] \]

3. **Substitute the given values:**
- Given:
- \(Λ^∞ [Mg^{2+}] = 53.06 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)
- \(Λ^∞ [SO₄^{2-}] = 80 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)
- \(Λ^∞ [Al^{3+}] = 63 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)

4. **Calculate Λ_m^∞ for Al₂(SO₄)₃:**
\[ Λ_m^∞ [Al₂(SO₄)₃] = 2 \cdot 63 + 3 \cdot 80 \]
\[ = 126 + 240 = 366 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

5. **Calculate Λ_m^∞ for MgSO₄:**
\[ Λ_m^∞ [MgSO₄] = 1 \cdot 53.06 + 1 \cdot 80 \]
\[ = 53.06 + 80 = 133.06 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

### Final Results:
- \(Λ_m^∞ [Al₂(SO₄)₃] = 366 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)
- \(Λ_m^∞ [MgSO₄] = 133.06 \, \text{ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\)