`^^_(m_((NH_(4)OH)))^(o)` is equal to
(a)`Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_((HCl))^(@)`
(b)`Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_((NaCl))^(@)`
(c)`Lambda_(m(NH_(4)Cl))^(@)+Lambda_(m(NaCl))^(@)-Lambda_((NaOH))^(@)`
(d)`Lambda_(m(NaOH))^(@)+Lambda_(m(NaOH))^(@)+Lambda_(m(NaCl))^(@)-Lambda_((NH_(4)Cl))^(@)`

Options

A. `Lambda_(m)^(@)(NH_(4)OH)+Lambda_(m)^(@)(NH_(4)Cl)-Lambda_((HCl))^(@)`
B. `Lambda_(m)^(@)(NH_(4)Cl)+Lambda_(m)^(@)(NaOH)-Lambda_((NaCl))^(@)`
C. `Lambda_(m(NH_(4)Cl))^(@)+Lambda_(m(NaCl))^(@)-Lambda_((NaOH))^(@)`
D. `Lambda_(m(NaOH))^(@)+Lambda_(m(NaOH))^(@)+Lambda_(m(NaCl))^(@)-Lambda_((NH_(4)Cl))^(@)`

Explanation

To solve the problem, we need to find the expression for the molar conductivity at infinite dilution for ammonium hydroxide (NH₄OH) using Kohlrausch's law.

### Step-by-Step Solution:

1. **Understanding Kohlrausch's Law**:
According to Kohlrausch's law, the molar conductivity at infinite dilution (Λₘ) of an electrolyte can be expressed as the sum of the molar conductivities of its individual ions.
\[
\Lambda_{m}^{\infty}(NH_4OH) = \Lambda_{m}^{\infty}(NH_4^+) + \Lambda_{m}^{\infty}(OH^-)
...