The value of `mu^(prop)` for `NH_(4)Cl, NaOH` and `NaCl` are `129.8, 248.1` and `126.4 ohm^(-1) cm^(2) mol^(-1)` respectively. Calculate `mu^(prop)` for `NH_(3)OH` solution.

Explanation

To calculate the molar conductivity at infinite dilution (\( \mu^{prop} \)) for \( NH_3OH \), we will use the values provided for \( NH_4Cl \), \( NaOH \), and \( NaCl \). The relationship we will use is based on the principle of additivity of molar conductivities.

### Step-by-Step Solution:

1. **Identify the given values:**
- \( \mu^{prop}(NH_4Cl) = 129.8 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( \mu^{prop}(NaOH) = 248.1 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
- \( \mu^{prop}(NaCl) = 126.4 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
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