From the following molar conductivities at infinite dilution `:`
`wedge_(m)^(@)` for `Ba(OH)_(2)=457.6Omega^(1)cm^(2)mol^(-1)`
`wedge_(m)^(@)` for `BaCl_(2)=240.6Omega^(-1) cm^(2)mol^(-1)`
`wedge_(m)^(@)` for `NH_(4)Cl=129.8 Omega^(-1) cm^(2) mol^(-1)`
Calculate `wedge_(m)^(@)` for `NH_(4)OH`.

Explanation

To calculate the molar conductivity at infinite dilution for \( \text{NH}_4\text{OH} \), we can use the given molar conductivities of the other compounds and apply the principle of superposition of ionic conductivities. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Molar Conductivities**:
- \( \Lambda_m^\circ \) for \( \text{Ba(OH)}_2 = 457.6 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- \( \Lambda_m^\circ \) for \( \text{BaCl}_2 = 240.6 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
- \( \Lambda_m^\circ \) for \( \text{NH}_4\text{Cl} = 129.8 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1} \)
...