Given that `wedge_(m)^(@) = 133.45 cm^(2) mol^(-1) (AgNO_(3))` `wedge_(m)^(@) = 149.9 5 cm^(2) mol^(-1)` (KCl), `wedge_(m)^(@) = 144.0`
`(KNO_(3))` the molar conductivity of AgCl at infinite diltue

Options

A. `132 S cm^(2)"mol"^(-1)`
B. `140 S cm^(2) mol^(-1)`
C. `138 S cm^(2) mol^(-1)`
D. `134 S cm^(2) mol^(-1)`

Explanation

To find the molar conductivity of AgCl at infinite dilution using the given data, we will apply Kohlrausch's law of independent migration of ions. Here’s a step-by-step solution:

### Step 1: Write down the given values
We have the following molar conductivities at infinite dilution:
- Molar conductivity of AgNO3, \( \Lambda_m^{\circ} (AgNO_3) = 133.45 \, \text{cm}^2 \, \text{mol}^{-1} \)
- Molar conductivity of KCl, \( \Lambda_m^{\circ} (KCl) = 149.95 \, \text{cm}^2 \, \text{mol}^{-1} \)
- Molar conductivity of KNO3, \( \Lambda_m^{\circ} (KNO_3) = 144.0 \, \text{cm}^2 \, \text{mol}^{-1} \)

### Step 2: Apply Kohlrausch's Law
According to Kohlrausch's law:
\[ \Lambda_m^{\circ} (AgCl) = \Lambda_m^{\circ} (AgNO_3) + \Lambda_m^{\circ} (KCl) - \Lambda_m^{\circ} (KNO_3) \]

### Step 3: Substitute the values into the equation
Now, we will substitute the given values into the equation:
\[ \Lambda_m^{\circ} (AgCl) = 133.45 + 149.95 - 144.0 \]

### Step 4: Perform the calculation
Now, we will perform the arithmetic:
\[ \Lambda_m^{\circ} (AgCl) = 133.45 + 149.95 - 144.0 = 139.4 \, \text{cm}^2 \, \text{mol}^{-1} \]

### Step 5: Conclusion
Thus, the molar conductivity of AgCl at infinite dilution is:
\[ \Lambda_m^{\circ} (AgCl) = 139.4 \, \text{cm}^2 \, \text{mol}^{-1} \]