Given molar conductivity of an infinite dilution: `wedge_(m)^(@)` for `Ba(OH)_(2)=517.6Omega^(-1)cm^(2)mol^(-1)`.
`wedge_(m)^(@)` for `BaCl_(2)=240.6Omega^(-1)cm^(2)mol^(-1),wedge_(m)^(@)` for `NH_(4)Cl=129.8Omega^(-1)cm^(2)mol^(-1)`. Calculate `wedge_(m)^(@)` for `NH_(4)OH`.

Explanation

To calculate the molar conductivity of ammonium hydroxide (NH₄OH), we can use the provided molar conductivities of the other compounds and apply the concept of ionic conductivities. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the expression for molar conductivity of NH₄OH
The molar conductivity of NH₄OH can be expressed in terms of its ions:
\[ \Lambda_m^{\circ} (NH_4OH) = \Lambda_m^{\circ} (NH_4^+) + \Lambda_m^{\circ} (OH^-) \]

### Step 2: Use the given data for Ba(OH)₂
For barium hydroxide, Ba(OH)₂, we can express its molar conductivity as:
\[ \Lambda_m^{\circ} (Ba(OH)_2) = \Lambda_m^{\circ} (Ba^{2+}) + 2 \Lambda_m^{\circ} (OH^-) \]
Given:
\[ \Lambda_m^{\circ} (Ba(OH)_2) = 517.6 \, \Omega^{-1} cm^2 mol^{-1} \]

### Step 3: Use the given data for BaCl₂
For barium chloride, BaCl₂, we can express its molar conductivity as:
\[ \Lambda_m^{\circ} (BaCl_2) = \Lambda_m^{\circ} (Ba^{2+}) + 2 \Lambda_m^{\circ} (Cl^-) \]
Given:
\[ \Lambda_m^{\circ} (BaCl_2) = 240.6 \, \Omega^{-1} cm^2 mol^{-1} \]

### Step 4: Use the given data for NH₄Cl
For ammonium chloride, NH₄Cl, we can express its molar conductivity as:
\[ \Lambda_m^{\circ} (NH_4Cl) = \Lambda_m^{\circ} (NH_4^+) + \Lambda_m^{\circ} (Cl^-) \]
Given:
\[ \Lambda_m^{\circ} (NH_4Cl) = 129.8 \, \Omega^{-1} cm^2 mol^{-1} \]

### Step 5: Set up the equations
From the equations above, we can derive the following:
1. From Ba(OH)₂:
\[ \Lambda_m^{\circ} (Ba^{2+}) + 2 \Lambda_m^{\circ} (OH^-) = 517.6 \]
2. From BaCl₂:
\[ \Lambda_m^{\circ} (Ba^{2+}) + 2 \Lambda_m^{\circ} (Cl^-) = 240.6 \]
3. From NH₄Cl:
\[ \Lambda_m^{\circ} (NH_4^+) + \Lambda_m^{\circ} (Cl^-) = 129.8 \]

### Step 6: Solve for ionic conductivities
We can manipulate these equations to isolate the variables. First, we can express \( \Lambda_m^{\circ} (Cl^-) \) in terms of \( \Lambda_m^{\circ} (NH_4^+) \):
\[ \Lambda_m^{\circ} (Cl^-) = 129.8 - \Lambda_m^{\circ} (NH_4^+) \]

### Step 7: Substitute and solve
Substituting this into the equation for BaCl₂:
\[ \Lambda_m^{\circ} (Ba^{2+}) + 2(129.8 - \Lambda_m^{\circ} (NH_4^+)) = 240.6 \]
This simplifies to:
\[ \Lambda_m^{\circ} (Ba^{2+}) + 259.6 - 2\Lambda_m^{\circ} (NH_4^+) = 240.6 \]
\[ \Lambda_m^{\circ} (Ba^{2+}) - 2\Lambda_m^{\circ} (NH_4^+) = -19 \]

### Step 8: Use Ba(OH)₂ equation
Now, we can use the Ba(OH)₂ equation:
\[ \Lambda_m^{\circ} (Ba^{2+}) + 2\Lambda_m^{\circ} (OH^-) = 517.6 \]
We can express \( \Lambda_m^{\circ} (OH^-) \) in terms of \( \Lambda_m^{\circ} (Ba^{2+}) \):
\[ \Lambda_m^{\circ} (OH^-) = \frac{517.6 - \Lambda_m^{\circ} (Ba^{2+})}{2} \]

### Step 9: Substitute back to find \( \Lambda_m^{\circ} (NH_4OH) \)
Finally, we can substitute back to find \( \Lambda_m^{\circ} (NH_4OH) \):
\[ \Lambda_m^{\circ} (NH_4OH) = \Lambda_m^{\circ} (NH_4^+) + \Lambda_m^{\circ} (OH^-) \]

### Step 10: Calculate the final value
After substituting the values and solving, we find:
\[ \Lambda_m^{\circ} (NH_4OH) = 268.3 \, \Omega^{-1} cm^2 mol^{-1} \]

### Final Answer:
\[ \Lambda_m^{\circ} (NH_4OH) = 268.3 \, \Omega^{-1} cm^2 mol^{-1} \]