The value of `Lambda_(m)^(@)" for "NH_(4) Cl, NaOH and NaCl" are 129.8, 248.1 and 126.4 Ohm"^(-1)" cm"^(2)" mol"^(-1)` respectively. Calculate `Lambda_(m)^(@)` for `NH_(4)OH` solution.

Options

A. 215.5
B. 251.5
C. 244.7
D. 351.5

Answer

To calculate the molar conductivity (Λ_m^0) of NH₄OH using the given values for NH₄Cl, NaOH, and NaCl, we will apply Kohlrausch's law. Here’s the step-by-step solution:

### Step 1: Understand the Components
We have the following molar conductivities:
- Λ_m^0 (NH₄Cl) = 129.8 Ohm^(-1) cm^2 mol^(-1)
- Λ_m^0 (NaOH) = 248.1 Ohm^(-1) cm^2 mol^(-1)
- Λ_m^0 (NaCl) = 126.4 Ohm^(-1) cm^2 mol^(-1)

### Step 2: Write the Equation Using Kohlrausch's Law
According to Kohlrausch's law, the molar conductivity of a compound can be calculated by adding the molar conductivities of its constituent ions. For NH₄OH, we can express it as:
\[ Λ_m^0 (NH₄OH) = Λ_m^0 (NH₄Cl) + Λ_m^0 (NaOH) - Λ_m^0 (NaCl) \]

### Step 3: Substitute the Values
Now, substitute the values into the equation:
\[ Λ_m^0 (NH₄OH) = 129.8 + 248.1 - 126.4 \]

### Step 4: Perform the Calculation
Now, perform the arithmetic:
1. Add 129.8 and 248.1:
\[ 129.8 + 248.1 = 377.9 \]
2. Subtract 126.4 from 377.9:
\[ 377.9 - 126.4 = 251.5 \]

### Step 5: Conclusion
Thus, the molar conductivity of NH₄OH is:
\[ Λ_m^0 (NH₄OH) = 251.5 \, \text{Ohm}^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \]

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