Molar conductivities at infinite dilution of `Mg^(2+)` and `Br^(-)` are 105.8 `Omega^(-1) cm^(-2) mol^(-1) ` and `78.2 Omega^(-1) cm^(-2) mol^(-1) ` respectively . Calculate molar conductivity at zero concentration of `MgBr_(2)`

Explanation

To calculate the molar conductivity at zero concentration (or at infinite dilution) of magnesium bromide (MgBr₂), we will use the molar conductivities of the individual ions, Mg²⁺ and Br⁻.

### Step-by-Step Solution:

1. **Identify the Molar Conductivities**:
- Molar conductivity of Mg²⁺, \( \Lambda^0_{Mg^{2+}} = 105.8 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- Molar conductivity of Br⁻, \( \Lambda^0_{Br^{-}} = 78.2 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)

2. **Determine the Molar Conductivity of MgBr₂**:
- The formula for the molar conductivity of an electrolyte is given by:
\[ \Lambda^0_{MgBr_2} = \Lambda^0_{Mg^{2+}} + 2 \cdot \Lambda^0_{Br^{-}} \]
- Here, we multiply the molar conductivity of Br⁻ by 2 because there are two bromide ions for every magnesium ion in the formula of magnesium bromide.

3. **Substitute the Values**:
- Substitute the values into the equation:
\[ \Lambda^0_{MgBr_2} = 105.8 + 2 \cdot 78.2 \]

4. **Calculate**:
- First, calculate \( 2 \cdot 78.2 \):
\[ 2 \cdot 78.2 = 156.4 \]
- Now, add this to the molar conductivity of Mg²⁺:
\[ \Lambda^0_{MgBr_2} = 105.8 + 156.4 = 262.2 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

5. **Final Result**:
- The molar conductivity at zero concentration of MgBr₂ is:
\[ \Lambda^0_{MgBr_2} = 262.2 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]