Calculate the molar conductivity at infinite dilution of AgCl from the following data:
`Lambda^@ (Ag NO_3) = 13.34 m S m^(2) mol^(-1) , Lambda^(@) (KCl) = 14.99 m S m^(2) mol^(-1)` and `Lambda^(@) (KNO_(3)) = 14.40 m S m^(2) mol^(-1)`

Explanation

To calculate the molar conductivity at infinite dilution of AgCl (\( \Lambda^0_{AgCl} \)), we can use the provided molar conductivities of AgNO3, KCl, and KNO3. The relationship we can use is based on the fact that the molar conductivity of a salt at infinite dilution is the sum of the molar conductivities of its constituent ions.

### Step-by-step Solution:

1. **Write the equation for molar conductivity of AgCl**:
\[ \Lambda^0_{AgCl} = \Lambda^0_{Ag^+} + \Lambda^0_{Cl^-} \]

2. **Express the molar conductivities of the known salts**:
- For AgNO3:
\[ \Lambda^0_{AgNO3} = \Lambda^0_{Ag^+} + \Lambda^0_{NO_3^-} = 13.34 \, \text{mS m}^2 \text{mol}^{-1} \]
- For KCl:
\[ \Lambda^0_{KCl} = \Lambda^0_{K^+} + \Lambda^0_{Cl^-} = 14.99 \, \text{mS m}^2 \text{mol}^{-1} \]
- For KNO3:
\[ \Lambda^0_{KNO3} = \Lambda^0_{K^+} + \Lambda^0_{NO_3^-} = 14.40 \, \text{mS m}^2 \text{mol}^{-1} \]

3. **Set up the equation to find \( \Lambda^0_{AgCl} \)**:
To find \( \Lambda^0_{AgCl} \), we can rearrange the equations:
\[ \Lambda^0_{AgCl} = \Lambda^0_{AgNO3} + \Lambda^0_{KCl} - \Lambda^0_{KNO3} \]

4. **Substitute the known values**:
\[ \Lambda^0_{AgCl} = 13.34 \, \text{mS m}^2 \text{mol}^{-1} + 14.99 \, \text{mS m}^2 \text{mol}^{-1} - 14.40 \, \text{mS m}^2 \text{mol}^{-1} \]

5. **Perform the calculation**:
\[ \Lambda^0_{AgCl} = 13.34 + 14.99 - 14.40 \]
\[ \Lambda^0_{AgCl} = 13.34 + 0.59 = 13.93 \, \text{mS m}^2 \text{mol}^{-1} \]

### Final Result:
The molar conductivity at infinite dilution of AgCl is:
\[ \Lambda^0_{AgCl} = 13.93 \, \text{mS m}^2 \text{mol}^{-1} \]