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. `139.5"ohm"^(-1) "cm"^(2) "eq"^(-1)`
B. `203"ohm"^(-1) "cm"^(2) "eq"^(-1)`
C. `279"ohm"^(-1) "cm"^(2) "eq"^(-1)`
D. `101.5"ohm"^(-1) "cm"^(2) "eq"^(-1)`

Explanation

To find the equivalent conductance of BaCl₂ at infinite dilution, we can follow these steps:

### Step 1: Identify the given values
- Molar conductance of \( Ba^{2+} \) = 127 \( \Omega^{-1} \, cm^{-1} \, mol^{-1} \)
- Molar conductance of \( Cl^{-} \) = 76 \( \Omega^{-1} \, cm^{-1} \, mol^{-1} \)

### Step 2: Calculate the equivalent conductance of each ion
- The equivalent conductance (\( \Lambda_{eq} \)) of \( Ba^{2+} \) is calculated by dividing the molar conductance by its valency (2):
\[ \Lambda_{eq}(Ba^{2+}) = \frac{127}{2} = 63.5 \, \Omega^{-1} \, cm^{-1} \, eq^{-1} \]

- The equivalent conductance of \( Cl^{-} \) is calculated by dividing the molar conductance by its valency (1):
\[ \Lambda_{eq}(Cl^{-}) = \frac{76}{1} = 76 \, \Omega^{-1} \, cm^{-1} \, eq^{-1} \]

### Step 3: Calculate the total equivalent conductance of \( BaCl_2 \)
- The equivalent conductance of \( BaCl_2 \) at infinite dilution is the sum of the equivalent conductance of its constituent ions:
\[ \Lambda_{eq}(BaCl_2) = \Lambda_{eq}(Ba^{2+}) + \Lambda_{eq}(Cl^{-}) \]
\[ \Lambda_{eq}(BaCl_2) = 63.5 + 76 = 139.5 \, \Omega^{-1} \, cm^{-1} \, eq^{-1} \]

### Final Answer
The equivalent conductance of \( BaCl_2 \) at infinite dilution is \( 139.5 \, \Omega^{-1} \, cm^{-1} \, eq^{-1} \).

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