`lambda_m(NH_4OH)^0` is equal to …..A….here A refers to

Options

A. `lambda_m(NH_4OH)^0+lambda_m(NH_4Cl)^0-lambda_(HCl)^0`
B. `lambda_m(NH_4Cl)^0+lambda_m(NaOH)^0-lambda_(NaCl)^0`
C. `lambda_m(NH_4Cl)^0+lambda_m(NaCl)^0-lambda_(NaOH)^0`
D. `lambda_m(NaOH)^0+lambda_m(NaCl)^0-lambda_(NH_4Cl)^0`

Answer

To find the limiting molar conductivity of ammonium hydroxide (NH₄OH), we can follow these steps:

### Step 1: Understand the concept of limiting molar conductivity
Limiting molar conductivity (\( \lambda_m^0 \)) is defined as the conductivity of an electrolyte in a very dilute solution, where the ions are far enough apart that they do not interact with each other. It is the sum of the contributions from each ion produced when the electrolyte dissociates.

### Step 2: Identify the ions produced by NH₄OH
Ammonium hydroxide (NH₄OH) dissociates in solution to produce ammonium ions (NH₄⁺) and hydroxide ions (OH⁻):
\[ \text{NH}_4\text{OH} \rightarrow \text{NH}_4^+ + \text{OH}^- \]

### Step 3: Write the expression for limiting molar conductivity
The limiting molar conductivity of NH₄OH can be expressed as the sum of the limiting molar conductivities of its ions:
\[ \lambda_m^0(\text{NH}_4\text{OH}) = \lambda_m^0(\text{NH}_4^+) + \lambda_m^0(\text{OH}^-) \]

### Step 4: Conclusion
Thus, the limiting molar conductivity of ammonium hydroxide (\( \lambda_m^0(\text{NH}_4\text{OH}) \)) is equal to the sum of the limiting molar conductivities of the ammonium ion and the hydroxide ion.

### Final Answer
So, \( A \) refers to \( \lambda_m^0(\text{NH}_4^+) + \lambda_m^0(\text{OH}^-) \).

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