The molar conductivity of `NH_(4)Cl` at infinite dilution is `149.7" S "cm^(2)mol^(-1)` and the ionic conductivities of `OH^(-)` and `Cl^(-)` and are 198 and 96.3 S `cm^(2)mol^(-1)` respectively. Calculate the molar conductivity of `NH_(4)OH` at this dilution.

Explanation

To calculate the molar conductivity of `NH₄OH` at infinite dilution, we can use the relationship between the molar conductivities of the ions and the overall molar conductivity of the compound.

### Step-by-Step Solution:

1. **Identify the given data:**
- Molar conductivity of `NH₄Cl` at infinite dilution, \( \Lambda^0_{NH₄Cl} = 149.7 \, \text{S cm}^2 \text{mol}^{-1} \)
- Ionic conductivity of `Cl⁻`, \( \Lambda^0_{Cl^-} = 96.3 \, \text{S cm}^2 \text{mol}^{-1} \)
- Ionic conductivity of `OH⁻`, \( \Lambda^0_{OH^-} = 198 \, \text{S cm}^2 \text{mol}^{-1} \)

2. **Write the formula for the molar conductivity of `NH₄OH`:**
The molar conductivity of a salt at infinite dilution can be expressed as the sum of the molar conductivities of its constituent ions:
\[ \Lambda^0_{NH₄OH} = \Lambda^0_{NH₄^+} + \Lambda^0_{OH^-} \]

3. **Rearrange to find \( \Lambda^0_{NH₄^+} \):**
We can find the molar conductivity of the ammonium ion by rearranging the equation:
\[ \Lambda^0_{NH₄^+} = \Lambda^0_{NH₄Cl} - \Lambda^0_{Cl^-} \]

4. **Substitute the known values:**
\[ \Lambda^0_{NH₄^+} = 149.7 \, \text{S cm}^2 \text{mol}^{-1} - 96.3 \, \text{S cm}^2 \text{mol}^{-1} \]
\[ \Lambda^0_{NH₄^+} = 53.4 \, \text{S cm}^2 \text{mol}^{-1} \]

5. **Calculate the molar conductivity of `NH₄OH`:**
Now substitute \( \Lambda^0_{NH₄^+} \) and \( \Lambda^0_{OH^-} \) into the equation for \( \Lambda^0_{NH₄OH} \):
\[ \Lambda^0_{NH₄OH} = 53.4 \, \text{S cm}^2 \text{mol}^{-1} + 198 \, \text{S cm}^2 \text{mol}^{-1} \]
\[ \Lambda^0_{NH₄OH} = 251.4 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
The molar conductivity of `NH₄OH` at infinite dilution is \( 251.4 \, \text{S cm}^2 \text{mol}^{-1} \).