Limiting molar conductivity of `NH_(4)OH` [i.e., `Lambda_(m)^(@)(NH_(4)OH)`] is equal to:

Options

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

Explanation

To find the limiting molar conductivity of \( NH_4OH \) (i.e., \( \Lambda_m^{(@)}(NH_4OH) \)), we can use Kohlrausch's law. According to this law, the molar conductivity of an electrolyte can be expressed as the sum of the molar conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Identify the ions in \( NH_4OH \)**:
- \( NH_4OH \) dissociates into \( NH_4^+ \) (ammonium ion) and \( OH^- \) (hydroxide ion).

2. **Write the expression for molar conductivity**:
- According to Kohlrausch's law:
\[ \Lambda_m^{(@)}(NH_4OH) = \Lambda_m(NH_4^+) + \Lambda_m(OH^-) \]

3. **Gather information from given data**:
- We need the molar conductivities of \( NH_4Cl \), \( NaCl \), and \( NaOH \) to find \( \Lambda_m(NH_4^+) \) and \( \Lambda_m(OH^-) \).
- From the dissociation of these salts, we can write:
- For \( NH_4Cl \):
\[ \Lambda_m(NH_4Cl) = \Lambda_m(NH_4^+) + \Lambda_m(Cl^-) \]
- For \( NaCl \):
\[ \Lambda_m(NaCl) = \Lambda_m(Na^+) + \Lambda_m(Cl^-) \]
- For \( NaOH \):
\[ \Lambda_m(NaOH) = \Lambda_m(Na^+) + \Lambda_m(OH^-) \]

4. **Set up the equations**:
- Let’s denote:
- \( \Lambda_m(NH_4^+) = x \)
- \( \Lambda_m(OH^-) = y \)
- \( \Lambda_m(Na^+) \) and \( \Lambda_m(Cl^-) \) are known from the data.

5. **Combine the equations**:
- From the equations:
- \( x + \Lambda_m(Cl^-) = \Lambda_m(NH_4Cl) \) (1)
- \( \Lambda_m(Na^+) + \Lambda_m(Cl^-) = \Lambda_m(NaCl) \) (2)
- \( \Lambda_m(Na^+) + y = \Lambda_m(NaOH) \) (3)

6. **Solve for \( x \) and \( y \)**:
- From equations (1) and (2), we can eliminate \( \Lambda_m(Cl^-) \):
\[ \Lambda_m(NH_4Cl) - \Lambda_m(NaCl) = x - y \]
- Rearranging gives us:
\[ x + y = \Lambda_m(NH_4Cl) + \Lambda_m(NaOH) - \Lambda_m(NaCl) \]

7. **Final expression for \( \Lambda_m^{(@)}(NH_4OH) \)**:
- Substitute \( x \) and \( y \) into the equation for \( \Lambda_m^{(@)}(NH_4OH) \):
\[ \Lambda_m^{(@)}(NH_4OH) = x + y = \Lambda_m(NH_4Cl) + \Lambda_m(NaOH) - \Lambda_m(NaCl) \]

### Final Answer:
\[ \Lambda_m^{(@)}(NH_4OH) = \Lambda_m(NH_4Cl) + \Lambda_m(NaOH) - \Lambda_m(NaCl) \]