The molar conducatance of `Ba^(2+)` and `Cl^(-)` are `127` and `76 ohm^(-1) cm^(-1) mol^(-1)` respectively at infinite dilution. The equivalent conductance of `BaCl_(2)` at infinte dilution will be

Options

A. `139.52`
B. `203`
C. `279`
D. `101.5`

Explanation

To find the equivalent conductance of BaCl₂ at infinite dilution, we can use Kohlrausch's law, which states that the equivalent conductance of an electrolyte at infinite dilution is the sum of the contributions from its individual ions.

### Step-by-Step Solution:

1. **Identify the ions and their molar conductances**:
- The ions in BaCl₂ are Ba²⁺ and Cl⁻.
- Given:
- Molar conductance of Ba²⁺ (λ⁰_Ba²⁺) = 127 Ω⁻¹ cm⁻¹ mol⁻¹
- Molar conductance of Cl⁻ (λ⁰_Cl⁻) = 76 Ω⁻¹ cm⁻¹ mol⁻¹

2. **Determine the equivalent conductance of each ion**:
- The equivalent conductance (λ_eq) is calculated based on the charge of the ions:
- For Ba²⁺ (which is divalent), the equivalent conductance is:
\[ \lambda_{eq, Ba^{2+}} = \frac{\lambda^0_{Ba^{2+}}}{2} = \frac{127}{2} = 63.5 \, \text{Ω}^{-1} \text{cm}^{-1} \text{equivalent}^{-1} \]
- For Cl⁻ (which is monovalent), the equivalent conductance is:
\[ \lambda_{eq, Cl^{-}} = \lambda^0_{Cl^{-}} = 76 \, \text{Ω}^{-1} \text{cm}^{-1} \text{equivalent}^{-1} \]

3. **Calculate the total equivalent conductance for BaCl₂**:
- Using Kohlrausch's law:
\[ \lambda_{eq, BaCl_2} = \lambda_{eq, Ba^{2+}} + 2 \times \lambda_{eq, Cl^{-}} \]
- Since there are two Cl⁻ ions in BaCl₂:
\[ \lambda_{eq, BaCl_2} = 63.5 + 2 \times 76 \]
- Calculate:
\[ \lambda_{eq, BaCl_2} = 63.5 + 152 = 215.5 \, \text{Ω}^{-1} \text{cm}^{-1} \text{equivalent}^{-1} \]

4. **Final Result**:
- The equivalent conductance of BaCl₂ at infinite dilution is:
\[ \lambda_{eq, BaCl_2} = 215.5 \, \text{Ω}^{-1} \text{cm}^{-1} \text{equivalent}^{-1} \]