The specific conductivity of a solution containing 5g of anhydrous `BaCl_(2)` (mol.wt. = 208) in 1000 `cm^(3)` of a solution is found to be 0.0058 `"ohm"^(-1) cm^(-1)`. Calculate the molar and equivalent conductivity of the solution.

Explanation

To solve the problem, we need to calculate the molar conductivity and equivalent conductivity of the solution containing 5g of anhydrous BaCl₂ in 1000 cm³ of solution, given that the specific conductivity (κ) is 0.0058 ohm⁻¹ cm⁻¹.

### Step 1: Calculate the Molarity of the Solution

1. **Find the number of moles of BaCl₂**:
\[ \text{Moles of BaCl₂} = \frac{\text{Weight of BaCl₂}}{\text{Molecular Weight of BaCl₂}} = \frac{5 \text{ g}}{208 \text{ g/mol}} \approx 0.02404 \text{ mol} \]
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