The molar conductance of ammonium hydroxide solution of concentration 0.1 M, 0.01M and 0.001 M are 3.6, 11.3 and `34.0 ohm^(-1)cm^(-2)mol^(-1)` respectively. Calculate the degree of dissociation of `NH_(4)OH` at these concentrations. Molar conductance at infinite dilution for `NH_(4)OH` is 271.1 `ohm^(-1)cm^(-1)mol^(-1)`.

Explanation

To calculate the degree of dissociation (α) of ammonium hydroxide (NH₄OH) at different concentrations, we can use the formula:

\[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \]

where:
- \(\Lambda_m\) is the molar conductance at the given concentration.
- \(\Lambda_m^0\) is the molar conductance at infinite dilution.

Given data:
- Molar conductance at 0.1 M: \(\Lambda_m = 3.6 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)
- Molar conductance at 0.01 M: \(\Lambda_m = 11.3 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)
- Molar conductance at 0.001 M: \(\Lambda_m = 34.0 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)
- Molar conductance at infinite dilution: \(\Lambda_m^0 = 271.1 \, \Omega^{-1} \, cm^2 \, mol^{-1}\)

### Step 1: Calculate the degree of dissociation at 0.1 M
Using the formula:

\[ \alpha_{0.1} = \frac{\Lambda_{0.1}}{\Lambda^0} = \frac{3.6}{271.1} \]

Calculating:

\[ \alpha_{0.1} = \frac{3.6}{271.1} \approx 0.0133 \]

### Step 2: Calculate the degree of dissociation at 0.01 M
Using the formula:

\[ \alpha_{0.01} = \frac{\Lambda_{0.01}}{\Lambda^0} = \frac{11.3}{271.1} \]

Calculating:

\[ \alpha_{0.01} = \frac{11.3}{271.1} \approx 0.0417 \]

### Step 3: Calculate the degree of dissociation at 0.001 M
Using the formula:

\[ \alpha_{0.001} = \frac{\Lambda_{0.001}}{\Lambda^0} = \frac{34.0}{271.1} \]

Calculating:

\[ \alpha_{0.001} = \frac{34.0}{271.1} \approx 0.125 \]

### Summary of Results
- Degree of dissociation at 0.1 M: \(\alpha_{0.1} \approx 0.0133\)
- Degree of dissociation at 0.01 M: \(\alpha_{0.01} \approx 0.0417\)
- Degree of dissociation at 0.001 M: \(\alpha_{0.001} \approx 0.125\)