Molar conductivity of `NH_4OH` can be calculated by the equation.

Options

A. `Lambda_(NH_4OH)^@=Lambda_(Ba(OH)_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@`
B. `Lambda_(NH_4OH)^@=Lambda_(BaCl_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(Ba(OH)_2)^@`
C. `Lambda_(NH_4OH)^@=(Lambda_(Ba(OH)_2)^@+ 2Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@)/2`
D. `Lambda_(NH^4OH)^@=(Lambda_(NH_4Cl)^@+Lambda_(Ba(OH)_2)^@)/2`

Explanation

To calculate the molar conductivity of \( NH_4OH \) (ammonium hydroxide), we need to use the concept of limiting molar conductivity and the principles of Kohlrausch's law. Here’s a step-by-step solution:

### Step 1: Understand Limiting Molar Conductivity
Limiting molar conductivity (\( \Lambda^0 \)) of an electrolyte is defined as the conductivity of a solution divided by the molarity of the solution as the concentration approaches zero. For weak electrolytes like \( NH_4OH \), we can express its limiting molar conductivity in terms of its constituent ions.

### Step 2: Identify the Ions in \( NH_4OH \)
The dissociation of \( NH_4OH \) in water can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]
Thus, the limiting molar conductivity of \( NH_4OH \) can be expressed as:
\[ \Lambda^0_{NH_4OH} = \Lambda^0_{NH_4^+} + \Lambda^0_{OH^-} \]

### Step 3: Use Kohlrausch's Law
According to Kohlrausch's law, the limiting molar conductivity of a weak electrolyte can also be expressed in terms of the limiting molar conductivities of other related electrolytes. We will use the conductivities of \( Ba(OH)_2 \), \( NH_4Cl \), and \( BaCl_2 \) to derive the expression for \( NH_4OH \).

1. **For \( Ba(OH)_2 \)**:
\[ Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^- \]
Thus,
\[ \Lambda^0_{Ba(OH)_2} = \Lambda^0_{Ba^{2+}} + 2\Lambda^0_{OH^-} \]

2. **For \( NH_4Cl \)**:
\[ NH_4Cl \rightarrow NH_4^+ + Cl^- \]
Thus,
\[ \Lambda^0_{NH_4Cl} = \Lambda^0_{NH_4^+} + \Lambda^0_{Cl^-} \]

3. **For \( BaCl_2 \)**:
\[ BaCl_2 \rightarrow Ba^{2+} + 2Cl^- \]
Thus,
\[ \Lambda^0_{BaCl_2} = \Lambda^0_{Ba^{2+}} + 2\Lambda^0_{Cl^-} \]

### Step 4: Set Up the Equation
To find \( \Lambda^0_{NH_4OH} \), we can use the relationships derived from the above equations. We can manipulate the equations to express \( \Lambda^0_{NH_4OH} \) in terms of the other conductivities.

Using the equations:
- \( \Lambda^0_{Ba(OH)_2} = \Lambda^0_{Ba^{2+}} + 2\Lambda^0_{OH^-} \)
- \( \Lambda^0_{NH_4Cl} = \Lambda^0_{NH_4^+} + \Lambda^0_{Cl^-} \)
- \( \Lambda^0_{BaCl_2} = \Lambda^0_{Ba^{2+}} + 2\Lambda^0_{Cl^-} \)

We can derive:
\[ \Lambda^0_{NH_4OH} = \Lambda^0_{Ba(OH)_2} + \Lambda^0_{NH_4Cl} - \Lambda^0_{BaCl_2} \]

### Step 5: Choose the Correct Option
Based on the derived equation, we can see that the correct option that represents the limiting molar conductivity of \( NH_4OH \) is:
\[ \Lambda^0_{NH_4OH} = \Lambda^0_{Ba(OH)_2} + \Lambda^0_{NH_4Cl} - \Lambda^0_{BaCl_2} \]
This corresponds to **Option 1**.

### Final Answer
Thus, the molar conductivity of \( NH_4OH \) can be calculated by the equation:
\[ \Lambda^0_{NH_4OH} = \Lambda^0_{Ba(OH)_2} + \Lambda^0_{NH_4Cl} - \Lambda^0_{BaCl_2} \]