Calculate the limiting molar conductivities of
a)magnesium sulphate and b) calcium chloride.
Limiting molar conductivity of `Mg^(2+), Ca^(2+), SO_(4)^(2-) and CI^(-)` are respectively `106, 119, 160 and 76.3 "s cm''^(2) mol^(-1)`.

Answer

To calculate the limiting molar conductivities of magnesium sulfate (MgSO₄) and calcium chloride (CaCl₂), we will use the given limiting molar conductivities of the respective ions.

### Step-by-Step Solution:

#### a) Limiting Molar Conductivity of Magnesium Sulfate (MgSO₄)

1. **Identify the ions produced by the dissociation of MgSO₄**:
- Magnesium sulfate dissociates into magnesium ions (Mg²⁺) and sulfate ions (SO₄²⁻).
- The reaction can be represented as:
\[ \text{MgSO}_4 \rightarrow \text{Mg}^{2+} + \text{SO}_4^{2-} \]

2. **Write the expression for the limiting molar conductivity (λ₀) of MgSO₄**:
- The limiting molar conductivity of MgSO₄ is the sum of the limiting molar conductivities of its ions:
\[ \lambda_0 (\text{MgSO}_4) = \lambda_0 (\text{Mg}^{2+}) + \lambda_0 (\text{SO}_4^{2-}) \]

3. **Substitute the given values**:
- From the problem, we have:
- \(\lambda_0 (\text{Mg}^{2+}) = 106 \, \text{s cm}^{-2} \text{mol}^{-1}\)
- \(\lambda_0 (\text{SO}_4^{2-}) = 160 \, \text{s cm}^{-2} \text{mol}^{-1}\)
- Therefore:
\[ \lambda_0 (\text{MgSO}_4) = 106 + 160 = 266 \, \text{s cm}^{-2} \text{mol}^{-1} \]

#### b) Limiting Molar Conductivity of Calcium Chloride (CaCl₂)

1. **Identify the ions produced by the dissociation of CaCl₂**:
- Calcium chloride dissociates into calcium ions (Ca²⁺) and chloride ions (Cl⁻).
- The reaction can be represented as:
\[ \text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2 \text{Cl}^- \]

2. **Write the expression for the limiting molar conductivity (λ₀) of CaCl₂**:
- The limiting molar conductivity of CaCl₂ is the sum of the limiting molar conductivities of its ions:
\[ \lambda_0 (\text{CaCl}_2) = \lambda_0 (\text{Ca}^{2+}) + 2 \cdot \lambda_0 (\text{Cl}^-) \]

3. **Substitute the given values**:
- From the problem, we have:
- \(\lambda_0 (\text{Ca}^{2+}) = 119 \, \text{s cm}^{-2} \text{mol}^{-1}\)
- \(\lambda_0 (\text{Cl}^-) = 76.3 \, \text{s cm}^{-2} \text{mol}^{-1}\)
- Therefore:
\[ \lambda_0 (\text{CaCl}_2) = 119 + 2 \cdot 76.3 = 119 + 152.6 = 271.6 \, \text{s cm}^{-2} \text{mol}^{-1} \]

### Final Results:
- The limiting molar conductivity of magnesium sulfate (MgSO₄) is **266 s cm⁻² mol⁻¹**.
- The limiting molar conductivity of calcium chloride (CaCl₂) is **271.6 s cm⁻² mol⁻¹**.