Find out the molar conductivity of an aqueous solution of `BaCl_(2)` at infinite dilution when ionic conductances of `Ba^(2+)` and `Cl^(-)` ion are 127.30 S `cm^(2)mol^(-1)` and 76.34 S `cm^(2)mol^(-1)` respectively.

Explanation

To find the molar conductivity of an aqueous solution of \( \text{BaCl}_2 \) at infinite dilution, we can use the following formula:

\[ \Lambda^0_m = \Lambda^0_{Ba^{2+}} + 2 \Lambda^0_{Cl^-} \]

Where:
- \( \Lambda^0_m \) is the molar conductivity of \( \text{BaCl}_2 \) at infinite dilution.
- \( \Lambda^0_{Ba^{2+}} \) is the ionic conductivity of the \( \text{Ba}^{2+} \) ion.
- \( \Lambda^0_{Cl^-} \) is the ionic conductivity of the \( \text{Cl}^- \) ion.

### Step-by-step Solution:

1. **Identify the given values:**
- Ionic conductance of \( \text{Ba}^{2+} \): \( \Lambda^0_{Ba^{2+}} = 127.30 \, \text{S cm}^2 \text{mol}^{-1} \)
- Ionic conductance of \( \text{Cl}^- \): \( \Lambda^0_{Cl^-} = 76.34 \, \text{S cm}^2 \text{mol}^{-1} \)

2. **Write the formula for molar conductivity of \( \text{BaCl}_2 \):**
\[ \Lambda^0_m = \Lambda^0_{Ba^{2+}} + 2 \Lambda^0_{Cl^-} \]

3. **Substitute the known values into the formula:**
\[ \Lambda^0_m = 127.30 + 2 \times 76.34 \]

4. **Calculate the contribution from \( \text{Cl}^- \):**
\[ 2 \times 76.34 = 152.68 \]

5. **Add the contributions together:**
\[ \Lambda^0_m = 127.30 + 152.68 = 279.98 \, \text{S cm}^2 \text{mol}^{-1} \]

6. **Final result:**
\[ \Lambda^0_m \approx 279.99 \, \text{S cm}^2 \text{mol}^{-1} \]

### Conclusion:
The molar conductivity of an aqueous solution of \( \text{BaCl}_2 \) at infinite dilution is approximately \( 279.99 \, \text{S cm}^2 \text{mol}^{-1} \).