The equivalent conductances at infinite dilution of `NH_(4)Br, KOH and KBr" at "25^(@)C " are "271.52 xx 10^(-4) , 266.5 xx 10^(-4)` and `1.51.66 xx 10^(-4)Omega^(-1)m^(2) ("g. eq")^(-1)` respectively. Calculate the degree of dissociation of 0.01 N `NH_(4)OH` at `25^(@)C` if the equivalent conductance of 0.01 N `NH_(4)OH` solution is `16.28 xx 10^(-4)Omega^(-1) m^(2) ("g eq")^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the equivalent conductance at infinite dilution of NH₄OH

According to Kohlrausch's law, the equivalent conductance at infinite dilution of NH₄OH can be calculated using the equivalent conductances of its constituent ions:

\[ \Lambda^\circ_{NH_4OH} = \Lambda^\circ_{NH_4^+} + \Lambda^\circ_{OH^-} \]

We can obtain \(\Lambda^\circ_{NH_4^+}\) and \(\Lambda^\circ_{OH^-}\) using the given data:

\[ \Lambda^\circ_{NH_4^+} = \Lambda^\circ_{NH_4Br} = 271.52 \times 10^{-4} \, \Omega^{-1} m^2 (g \, eq)^{-1} \]
\[ \Lambda^\circ_{OH^-} = \Lambda^\circ_{KOH} = 266.5 \times 10^{-4} \, \Omega^{-1} m^2 (g \, eq)^{-1} \]

Now, we can calculate \(\Lambda^\circ_{NH_4OH}\):

\[ \Lambda^\circ_{NH_4OH} = 271.52 \times 10^{-4} + 266.5 \times 10^{-4} - 1.51 \times 10^{-4} \]

Calculating this gives:

\[ \Lambda^\circ_{NH_4OH} = 271.52 + 266.5 - 1.51 = 536.51 \times 10^{-4} \, \Omega^{-1} m^2 (g \, eq)^{-1} \]

### Step 2: Calculate the degree of dissociation (\(\alpha\))

The degree of dissociation can be calculated using the formula:

\[ \alpha = \frac{\Lambda_{NH_4OH}}{\Lambda^\circ_{NH_4OH}} \]

Where:
- \(\Lambda_{NH_4OH} = 16.28 \times 10^{-4} \, \Omega^{-1} m^2 (g \, eq)^{-1}\) (given)
- \(\Lambda^\circ_{NH_4OH} = 536.51 \times 10^{-4} \, \Omega^{-1} m^2 (g \, eq)^{-1}\) (calculated)

Substituting the values:

\[ \alpha = \frac{16.28 \times 10^{-4}}{536.51 \times 10^{-4}} \]

Calculating this gives:

\[ \alpha = \frac{16.28}{536.51} \approx 0.0304 \]

### Step 3: Convert to percentage

To express the degree of dissociation as a percentage, we multiply by 100:

\[ \alpha \times 100 \approx 0.0304 \times 100 \approx 3.04\% \]

### Final Answer

The degree of dissociation of 0.01 N \(NH_4OH\) at 25°C is approximately **3.04%**.

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