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)^(@)(Na_(4)Cl)-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 `NH₄OH` (ammonium hydroxide), we will apply Kohlrausch's Law, which states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.

### Step-by-step Solution:

1. **Identify the ions in `NH₄OH`:**
- The dissociation of `NH₄OH` in water gives us the following ions:
\[ NH₄OH \rightarrow NH₄^+ + OH^- \]
- Therefore, the ions present are `NH₄^+` (ammonium ion) and `OH^-` (hydroxide ion).

2. **Apply Kohlrausch's Law:**
- According to Kohlrausch's Law, the limiting molar conductivity of `NH₄OH` can be expressed as:
\[ \Lambda_m^0(NH₄OH) = \Lambda_m^0(NH₄^+) + \Lambda_m^0(OH^-) \]
- Here, `\Lambda_m^0(NH₄^+)` is the limiting molar conductivity of the ammonium ion, and `\Lambda_m^0(OH^-)` is the limiting molar conductivity of the hydroxide ion.

3. **Find the limiting molar conductivities of the ions:**
- We need to know the values of `\Lambda_m^0(NH₄^+)` and `\Lambda_m^0(OH^-)`. These values are typically provided in tables of ionic conductivities:
- For example, let's assume:
- `\Lambda_m^0(NH₄^+) = 73.5 S cm²/mol`
- `\Lambda_m^0(OH^-) = 198.5 S cm²/mol`
- (Note: Actual values may vary; please refer to the latest data or textbooks.)

4. **Calculate the limiting molar conductivity of `NH₄OH`:**
- Substitute the values into the equation:
\[ \Lambda_m^0(NH₄OH) = 73.5 + 198.5 = 272.0 S cm²/mol \]

5. **Conclusion:**
- The limiting molar conductivity of `NH₄OH` is:
\[ \Lambda_m^0(NH₄OH) = 272.0 S cm²/mol \]