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. cannot be predicted from given data

Explanation

To find the equivalent conductivity at infinite dilution (λ_eq^∞) for NH₄OH, we can use the following relationship:

\[ \lambda_{eq}^{\infty} (NH_4OH) = \lambda_{eq}^{\infty} (NH_4Cl) + \lambda_{eq}^{\infty} (NaOH) - \lambda_{eq}^{\infty} (NaCl) \]

### Step 1: Write down the known values.
- λ_eq^∞ (NH₄Cl) = 149.74 ohm⁻¹ cm² eq⁻¹
- λ_eq^∞ (NaOH) = 248.1 ohm⁻¹ cm² eq⁻¹
- λ_eq^∞ (NaCl) = 126.4 ohm⁻¹ cm² eq⁻¹

### Step 2: Substitute the known values into the equation.
\[ \lambda_{eq}^{\infty} (NH_4OH) = 149.74 + 248.1 - 126.4 \]

### Step 3: Perform the addition and subtraction.
1. First, add λ_eq^∞ (NH₄Cl) and λ_eq^∞ (NaOH):
\[ 149.74 + 248.1 = 397.84 \]
2. Then, subtract λ_eq^∞ (NaCl):
\[ 397.84 - 126.4 = 271.44 \]

### Step 4: Write down the final result.
\[ \lambda_{eq}^{\infty} (NH_4OH) = 271.44 \, \text{ohm}^{-1} \text{cm}^2 \text{eq}^{-1} \]

### Conclusion:
The value of λ_eq^∞ for NH₄OH is 271.44 ohm⁻¹ cm² eq⁻¹.

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