Calculate the equivalent conductance of `NH_4OH` at infinite dilution using Kohlrausch law. Given that `Lambda_(0)` values of `NaOH, NaCI and NH_4Cl` are respectively 217.4, 108.9 and `129"ohm"^(-1) cm^(2)`.

Answer

To calculate the equivalent conductance of `NH₄OH` at infinite dilution using Kohlrausch's law, we can follow these steps:

### Step 1: Understand Kohlrausch's Law
Kohlrausch's law states that the equivalent conductance at infinite dilution (Λ₀) of an electrolyte can be expressed as the sum of the conductances of its individual ions at infinite dilution. For `NH₄OH`, we can express it as:

\[ \Lambda_0(NH_4OH) = \Lambda_0(NH_4^+) + \Lambda_0(OH^-) \]

### Step 2: Relate `NH₄OH` to Known Electrolytes
We can relate the ions of `NH₄OH` to the known electrolytes `NH₄Cl` and `NaOH`. The relationship can be established using the following equation:

\[ \Lambda_0(NH_4OH) = \Lambda_0(NH_4^+) + \Lambda_0(OH^-) = \Lambda_0(NH_4Cl) - \Lambda_0(Cl^-) + \Lambda_0(NaOH) - \Lambda_0(Na^+) \]

### Step 3: Substitute Known Values
We have the following values:
- \(\Lambda_0(NH_4Cl) = 129 \, \Omega^{-1} cm^2\)
- \(\Lambda_0(NaOH) = 217.4 \, \Omega^{-1} cm^2\)
- \(\Lambda_0(NaCl) = 108.9 \, \Omega^{-1} cm^2\)

Using the relationship established in Step 2, we can substitute these values into the equation:

\[ \Lambda_0(NH_4OH) = \Lambda_0(NH_4Cl) - \Lambda_0(Cl^-) + \Lambda_0(NaOH) - \Lambda_0(Na^+) \]

### Step 4: Calculate the Equivalent Conductance
We know that \(\Lambda_0(Cl^-) = \Lambda_0(NaCl) - \Lambda_0(Na^+)\). Since \(\Lambda_0(Na^+) = \Lambda_0(Cl^-)\), we can simplify our equation:

\[ \Lambda_0(NH_4OH) = \Lambda_0(NH_4Cl) + \Lambda_0(NaOH) - \Lambda_0(NaCl) \]

Substituting the values:

\[ \Lambda_0(NH_4OH) = 129 + 217.4 - 108.9 \]

Calculating this gives:

\[ \Lambda_0(NH_4OH) = 237.5 \, \Omega^{-1} cm^2 \]

### Final Answer
The equivalent conductance of `NH₄OH` at infinite dilution is:

\[ \Lambda_0(NH_4OH) = 237.5 \, \Omega^{-1} cm^2 \]

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