The equivalent conductances of an infinitely dilute solution `NH_(4)CI` is 150 and the ionic conductances of `OH^(-)` and `CI^(-)` ions are 198 and 76 respectively. What will be the equivalent conductance of the solution of `NH_(4)OH` at infinite dilution. if the equivalent conductance of a `0.01N` solution `NH_(4)OH` is `9.6, what will be its degree of dissociation.

Explanation

To solve the problem step by step, we will break it down into two parts:

### Part 1: Calculate the Equivalent Conductance of NH₄OH at Infinite Dilution

1. **Identify the Known Values:**
- Equivalent conductance of NH₄Cl at infinite dilution (λ∞(NH₄Cl)) = 150 S cm²/equiv
- Ionic conductance of Cl⁻ (λ∞(Cl⁻)) = 76 S cm²/equiv
- Ionic conductance of OH⁻ (λ∞(OH⁻)) = 198 S cm²/equiv

2. **Calculate Ionic Conductance of NH₄⁺:**
- The equivalent conductance at infinite dilution for NH₄Cl can be expressed as:
\[ \lambda_{\infty} (NH₄Cl) = \lambda_{\infty} (NH₄^+) + \lambda_{\infty} (Cl^-) \]
- Rearranging gives:
\[ \lambda_{\infty} (NH₄^+) = \lambda_{\infty} (NH₄Cl) - \lambda_{\infty} (Cl^-) \]
- Substituting the known values:
\[ \lambda_{\infty} (NH₄^+) = 150 - 76 = 74 \, \text{S cm}^2/\text{equiv} \]

3. **Calculate Equivalent Conductance of NH₄OH at Infinite Dilution:**
- The equivalent conductance for NH₄OH at infinite dilution can be expressed as:
\[ \lambda_{\infty} (NH₄OH) = \lambda_{\infty} (NH₄^+) + \lambda_{\infty} (OH^-) \]
- Substituting the values we have:
\[ \lambda_{\infty} (NH₄OH) = 74 + 198 = 272 \, \text{S cm}^2/\text{equiv} \]

### Part 2: Calculate the Degree of Dissociation (α)

1. **Identify the Known Values:**
- Equivalent conductance of 0.01N solution of NH₄OH (λ) = 9.6 S cm²/equiv
- Equivalent conductance at infinite dilution (λ∞(NH₄OH)) = 272 S cm²/equiv

2. **Calculate the Degree of Dissociation (α):**
- The degree of dissociation can be calculated using the formula:
\[ \alpha = \frac{\lambda}{\lambda_{\infty}} \]
- Substituting the known values:
\[ \alpha = \frac{9.6}{272} \]
- Performing the calculation:
\[ \alpha = 0.0353 \]
- To express this as a percentage:
\[ \alpha \times 100 = 0.0353 \times 100 = 3.53\% \]

### Final Answers:
- Equivalent conductance of NH₄OH at infinite dilution = 272 S cm²/equiv
- Degree of dissociation (α) = 3.53%