`Lambda_((m)(NH_(4)OH))^(@)` is equal to

Options

A. `Lambda_(m(NH_4OH))^@ + Lambda_(m(NH_4Cl))^@-Lambda_(m(HCl))^@`
B. `Lambda_(m(NH_4Cl))^@+Lambda_(m(NaOH))^@-Lambda_(m(NaCl))^@`
C. `Lambda_(m(NH_4Cl))^@+Lambda_(m(NaCl))^@-Lambda_(m(NaOH))^@`
D. `Lambda_(m(NaOH))^@+Lambda_(m(NaCl))^@-Lambda_(m(NH_4Cl))^@`

Explanation

To find the molar conductivity at infinite dilution (\( \Lambda_m^\infty \)) of ammonium hydroxide (\( NH_4OH \)), we can use the principle that at infinite dilution, the molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions.

### Step-by-Step Solution:

1. **Understanding Molar Conductivity at Infinite Dilution**:
The molar conductivity at infinite dilution (\( \Lambda_m^\infty \)) of an electrolyte can be expressed as:
\[ \Lambda_m^\infty (NH_4OH) = \Lambda^\infty (NH_4^+) + \Lambda^\infty (OH^-) \]
where \( NH_4^+ \) is the ammonium ion and \( OH^- \) is the hydroxide ion.

2. **Using Known Values**:
We need the molar conductivities at infinite dilution for the relevant ions. We can derive \( \Lambda^\infty (NH_4OH) \) from the known values of other salts:
- For ammonium chloride (\( NH_4Cl \)):
\[ \Lambda^\infty (NH_4Cl) = \Lambda^\infty (NH_4^+) + \Lambda^\infty (Cl^-) \]
- For sodium hydroxide (\( NaOH \)):
\[ \Lambda^\infty (NaOH) = \Lambda^\infty (Na^+) + \Lambda^\infty (OH^-) \]
- For sodium chloride (\( NaCl \)):
\[ \Lambda^\infty (NaCl) = \Lambda^\infty (Na^+) + \Lambda^\infty (Cl^-) \]

3. **Setting Up the Equation**:
We can express \( \Lambda^\infty (NH_4OH) \) in terms of the known conductivities:
\[ \Lambda^\infty (NH_4OH) = \Lambda^\infty (NH_4Cl) + \Lambda^\infty (NaOH) - \Lambda^\infty (NaCl) \]

4. **Substituting the Values**:
Substitute the known values into the equation:
\[ \Lambda^\infty (NH_4OH) = \left(\Lambda^\infty (NH_4^+) + \Lambda^\infty (Cl^-)\right) + \left(\Lambda^\infty (Na^+) + \Lambda^\infty (OH^-)\right) - \left(\Lambda^\infty (Na^+) + \Lambda^\infty (Cl^-)\right) \]

5. **Simplifying the Equation**:
Upon simplifying, the terms \( \Lambda^\infty (Cl^-) \) and \( \Lambda^\infty (Na^+) \) cancel out:
\[ \Lambda^\infty (NH_4OH) = \Lambda^\infty (NH_4^+) + \Lambda^\infty (OH^-) \]

6. **Final Result**:
Thus, the molar conductivity at infinite dilution for ammonium hydroxide is:
\[ \Lambda_m^\infty (NH_4OH) = \Lambda^\infty (NH_4^+) + \Lambda^\infty (OH^-) \]