If a centi normal solution of `NH_4OH` has molar conductivity equal to `9.6 Omega^(-1) cm^(2) mol^(-1)`. what will be the per cent dissociation of `NH_4OH` at this dilution.

Answer

To solve the problem, we need to find the percent dissociation of ammonium hydroxide (NH₄OH) in a centi-normal solution given its molar conductivity. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Molar conductivity (Λ) of the centi-normal solution of NH₄OH = 9.6 Ω⁻¹ cm² mol⁻¹
- Molar conductivity at infinite dilution (Λ°) of NH₄OH = 240 Ω⁻¹ cm² mol⁻¹ (This value is generally known or can be looked up in tables.)

2. **Use the Formula for Percent Dissociation:**
- The formula for calculating the degree of dissociation (α) is given by:
\[ \alpha = \frac{\text{Molar Conductivity at given concentration (Λ)}}{\text{Molar Conductivity at infinite dilution (Λ°)}} \]

3. **Substitute the Values into the Formula:**
- Substitute the given values into the formula:
\[ \alpha = \frac{9.6 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}}{240 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}} \]

4. **Calculate α:**
- Perform the division:
\[ \alpha = \frac{9.6}{240} = 0.04 \]

5. **Convert α to Percent Dissociation:**
- To find the percent dissociation, multiply α by 100:
\[ \text{Percent Dissociation} = \alpha \times 100 = 0.04 \times 100 = 4\% \]

### Final Answer:
The percent dissociation of NH₄OH at this dilution is **4%**.