The specific conductivity of a solution containing `1.0g` of anhydrous `BaCl_(2)` in `200 cm^(3)` of the solution has been found to be `0.0058 S cm^(-1)`. Calculate the molar and equivalent conductivity of the solution. Molecular wt. of `BaCl_(2) = 208

Explanation

To solve the problem of calculating the molar and equivalent conductivity of a solution containing 1.0 g of anhydrous BaCl₂ in 200 cm³ of solution with a specific conductivity of 0.0058 S cm⁻¹, we will follow these steps:

### Step 1: Calculate Molarity of the Solution
Molarity (M) is defined as the number of moles of solute per liter of solution.

1. **Calculate the number of moles of BaCl₂:**
\[
\text{Moles of BaCl₂} = \frac{\text{Given weight}}{\text{Molecular weight}} = \frac{1.0 \, \text{g}}{208 \, \text{g/mol}} = \frac{1}{208} \, \text{mol}
...