The equivalent conductance of `Ba^(2+)` and `Cl^(-)` are `76ohm^(-1) cm^(2) eq^(-1)` and `63.5 ohm^(-1) cm^(2)eq^(-1)`, respectively, at infinite dilution . The equivalent conductance `(` in `ohm^(-1) cm^(2)eq^(-1))` of `BaCl_(2)` at infinite dilution will be

Options

A. 203
B. 279
C. 101.5
D. 139.5

Answer

To find the equivalent conductance of BaCl₂ at infinite dilution, we can use the following steps:

### Step 1: Understand the concept of equivalent conductance
Equivalent conductance (Λ) is a measure of the ability of an electrolyte to conduct electricity when dissolved in a solvent. At infinite dilution, each ion contributes to the total conductance of the solution.

### Step 2: Identify the ions in BaCl₂
Barium chloride (BaCl₂) dissociates into one barium ion (Ba²⁺) and two chloride ions (Cl⁻) in solution:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2 \text{Cl}^{-} \]

### Step 3: Write the formula for equivalent conductance of BaCl₂
The equivalent conductance of BaCl₂ at infinite dilution (Λ_BaCl₂) can be calculated using the equivalent conductance of its constituent ions:
\[ \Lambda_{\text{BaCl}_2} = \Lambda_{\text{Ba}^{2+}} + 2 \Lambda_{\text{Cl}^{-}} \]

### Step 4: Substitute the given values
From the problem, we know:
- \( \Lambda_{\text{Ba}^{2+}} = 76 \, \Omega^{-1} \, \text{cm}^2 \, \text{eq}^{-1} \)
- \( \Lambda_{\text{Cl}^{-}} = 63.5 \, \Omega^{-1} \, \text{cm}^2 \, \text{eq}^{-1} \)

Now substituting these values into the formula:
\[ \Lambda_{\text{BaCl}_2} = 76 + 2 \times 63.5 \]

### Step 5: Perform the calculations
Calculating the contribution from the chloride ions:
\[ 2 \times 63.5 = 127 \]

Now, add this to the conductance of barium:
\[ \Lambda_{\text{BaCl}_2} = 76 + 127 = 203 \, \Omega^{-1} \, \text{cm}^2 \, \text{eq}^{-1} \]

### Step 6: Conclusion
The equivalent conductance of BaCl₂ at infinite dilution is:
\[ \Lambda_{\text{BaCl}_2} = 203 \, \Omega^{-1} \, \text{cm}^2 \, \text{eq}^{-1} \]