The degree of association `(alpha)` is given by the expression

Options

A. `alpha=(i-1)/(1/n-1)`
B. `alpha=(i-1)/(n-1)`
C. `alpha=(1-i)/(1/n-1)`
D. `alpha=(1-i)/(n-1)`

Explanation

To find the degree of association \(\alpha\), we need to derive its expression step by step. Let's consider a scenario where a molecule \(A\) associates to form \(A_n\).

### Step-by-Step Solution:

1. **Initial Moles:**
- Assume initially we have 1 mole of \(A\).
- Therefore, initial moles of \(A_n\) is 0.

2. **At Equilibrium:**
- Let \(\alpha\) be the degree of association.
- At equilibrium, the moles of \(A\) will be \(1 - \alpha\).
- Since \(n\) moles of \(A\) associate to form 1 mole of \(A_n\), the moles of \(A_n\) at equilibrium will be \(\frac{\alpha}{n}\).

3. **Total Moles at Equilibrium:**
- The total moles at equilibrium will be the sum of moles of \(A\) and \(A_n\):
\[ \text{Total moles at equilibrium} = (1 - \alpha) + \frac{\alpha}{n} \]

4. **Van't Hoff Factor (i):**
- The Van't Hoff factor \(i\) is given by the ratio of the total moles at equilibrium to the initial moles:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} \]
\[ i = \frac{(1 - \alpha) + \frac{\alpha}{n}}{1} \]
\[ i = 1 - \alpha + \frac{\alpha}{n} \]

5. **Simplifying the Expression:**
- Rearrange the equation to solve for \(\alpha\):
\[ i = 1 - \alpha + \frac{\alpha}{n} \]
\[ i = 1 + \alpha \left(\frac{1}{n} - 1\right) \]
\[ i - 1 = \alpha \left(\frac{1}{n} - 1\right) \]

6. **Solving for \(\alpha\):**
- Isolate \(\alpha\):
\[ \alpha = \frac{i - 1}{\frac{1}{n} - 1} \]
\[ \alpha = \frac{i - 1}{\frac{1 - n}{n}} \]
\[ \alpha = \frac{(i - 1) \cdot n}{1 - n} \]

### Final Expression:
\[ \alpha = \frac{i - 1}{\frac{1}{n} - 1} \]