The molar conductivity of `NH_(4)Cl` at infinite dilution is 149.7`"ohm"^(-)cm^(2)"mol"^(-1)` and ionic conductances of hydroxyl ions and chloride ions are 198 and 76.3 respectively. Calculate the molar conductivity of `NH_(4)OH` at infinite dilution.

Explanation

To calculate the molar conductivity of \( NH_4OH \) at infinite dilution, we can use the relationship between the molar conductivities of the involved species. The relevant equation is:

\[ \Lambda_{m}(NH_4OH) = \Lambda_{m}(NH_4Cl) + \Lambda_{m}(OH^-) - \Lambda_{m}(Cl^-) \]

Where:
- \( \Lambda_{m}(NH_4OH) \) is the molar conductivity of ammonium hydroxide at infinite dilution.
- \( \Lambda_{m}(NH_4Cl) \) is the molar conductivity of ammonium chloride at infinite dilution, which is given as \( 149.7 \, \Omega^{-1} \, cm^2 \, mol^{-1} \).
- \( \Lambda_{m}(OH^-) \) is the ionic conductivity of hydroxyl ions, given as \( 198 \, \Omega^{-1} \, cm^2 \, mol^{-1} \).
- \( \Lambda_{m}(Cl^-) \) is the ionic conductivity of chloride ions, given as \( 76.3 \, \Omega^{-1} \, cm^2 \, mol^{-1} \).

Now, we can substitute the values into the equation:

1. Start with the known values:
- \( \Lambda_{m}(NH_4Cl) = 149.7 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( \Lambda_{m}(OH^-) = 198 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( \Lambda_{m}(Cl^-) = 76.3 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)

2. Substitute these values into the equation:
\[ \Lambda_{m}(NH_4OH) = 149.7 + 198 - 76.3 \]

3. Perform the calculations step by step:
- First, add \( 149.7 + 198 \):
\[ 149.7 + 198 = 347.7 \]
- Then subtract \( 76.3 \):
\[ 347.7 - 76.3 = 271.4 \]

4. Therefore, the molar conductivity of \( NH_4OH \) at infinite dilution is:
\[ \Lambda_{m}(NH_4OH) = 271.4 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

### Final Answer:
The molar conductivity of \( NH_4OH \) at infinite dilution is \( 271.4 \, \Omega^{-1} \, cm^2 \, mol^{-1} \).