The values of `wedge_(m)^(oo)` for`NH_(4)Cl,NaOH,` and `NaCl` are, respectively, `149.74,248.1`,and `126.4 ohm^(-1)cm^(2)eq^(-1)`. The value of `wedge_(eq)^(oo)NH_(4)OH` is

Options

A. `371.44`
B. `271.44`
C. `71.44`
D. It cannot be calculated from the data given.

Explanation

To find the value of molar conductivity at infinite dilution for ammonium hydroxide (NH₄OH), we can use the concept of molar conductivity and the principle of dilution as described by Kohlrausch's law. The molar conductivity at infinite dilution for a compound can be calculated by summing the contributions from its constituent ions.

### Step-by-Step Solution:

1. **Identify the Molar Conductivities:**
- Given values:
- \( \Lambda_m^\infty (NH_4Cl) = 149.74 \, \Omega^{-1} cm^2 eq^{-1} \)
- \( \Lambda_m^\infty (NaOH) = 248.1 \, \Omega^{-1} cm^2 eq^{-1} \)
- \( \Lambda_m^\infty (NaCl) = 126.4 \, \Omega^{-1} cm^2 eq^{-1} \)

2. **Write the Molar Conductivity Equations:**
- For \( NH_4Cl \):
\[ \Lambda_m^\infty (NH_4Cl) = \Lambda_m^\infty (NH_4^+) + \Lambda_m^\infty (Cl^-) \]
- For \( NaOH \):
\[ \Lambda_m^\infty (NaOH) = \Lambda_m^\infty (Na^+) + \Lambda_m^\infty (OH^-) \]
- For \( NaCl \):
\[ \Lambda_m^\infty (NaCl) = \Lambda_m^\infty (Na^+) + \Lambda_m^\infty (Cl^-) \]

3. **Set Up the Equations:**
- From the equations for \( NaCl \):
\[ \Lambda_m^\infty (Na^+) + \Lambda_m^\infty (Cl^-) = 126.4 \]
- Substitute \( \Lambda_m^\infty (Na^+) \) from \( NaOH \):
\[ \Lambda_m^\infty (Na^+) = \Lambda_m^\infty (NaOH) - \Lambda_m^\infty (OH^-) = 248.1 - \Lambda_m^\infty (OH^-) \]

4. **Combine the Equations:**
- Substitute \( \Lambda_m^\infty (Na^+) \) into the \( NaCl \) equation:
\[ (248.1 - \Lambda_m^\infty (OH^-)) + \Lambda_m^\infty (Cl^-) = 126.4 \]
- Rearranging gives:
\[ \Lambda_m^\infty (Cl^-) = 126.4 - 248.1 + \Lambda_m^\infty (OH^-) \]
- Thus:
\[ \Lambda_m^\infty (Cl^-) = \Lambda_m^\infty (OH^-) - 121.7 \]

5. **Substituting Back:**
- Now substitute \( \Lambda_m^\infty (Cl^-) \) back into the \( NH_4Cl \) equation:
\[ 149.74 = \Lambda_m^\infty (NH_4^+) + (\Lambda_m^\infty (OH^-) - 121.7) \]
- Rearranging gives:
\[ \Lambda_m^\infty (NH_4^+) = 149.74 - \Lambda_m^\infty (OH^-) + 121.7 \]
- Therefore:
\[ \Lambda_m^\infty (NH_4^+) = 271.44 - \Lambda_m^\infty (OH^-) \]

6. **Final Calculation:**
- Now, we can find \( \Lambda_m^\infty (NH_4OH) \):
\[ \Lambda_m^\infty (NH_4OH) = \Lambda_m^\infty (NH_4^+) + \Lambda_m^\infty (OH^-) \]
- Substitute the expression we found for \( \Lambda_m^\infty (NH_4^+) \):
\[ \Lambda_m^\infty (NH_4OH) = (271.44 - \Lambda_m^\infty (OH^-)) + \Lambda_m^\infty (OH^-) \]
- This simplifies to:
\[ \Lambda_m^\infty (NH_4OH) = 271.44 \, \Omega^{-1} cm^2 eq^{-1} \]

### Final Answer:
Thus, the value of \( \Lambda_{eq}^\infty (NH_4OH) \) is **271.44 \( \Omega^{-1} cm^2 eq^{-1} \)**.