The degree of dissociation `alpha` of a week electrolyte is
where `n` is the number of ions given by `1 mol` of electrolyte.

Options

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

Explanation

To find the degree of dissociation (α) of a weak electrolyte, we can follow these steps:

### Step 1: Understanding the Weak Electrolyte
Let’s consider a weak electrolyte represented as HA. When it dissociates in solution, it can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]

### Step 2: Defining Degree of Dissociation
The degree of dissociation (α) is defined as the fraction of the total number of moles of the electrolyte that dissociates into ions. If we start with 1 mole of HA, then at equilibrium, the amount of HA that has dissociated is α moles.

### Step 3: Determining the Number of Ions Produced
For a weak electrolyte like HA, when it dissociates, it produces \( n \) ions. In this case, for every mole of HA that dissociates, it produces 2 ions (H⁺ and A⁻), so \( n = 2 \).

### Step 4: Relating Van't Hoff Factor (i) to Degree of Dissociation
The Van't Hoff factor (i) is defined as the ratio of the number of particles in solution after dissociation to the number of formula units initially dissolved. For a weak electrolyte, the relationship can be expressed as:
\[ i = 1 + (n - 1)α \]
Where:
- \( i \) = Van't Hoff factor
- \( n \) = number of ions produced by 1 mole of electrolyte
- \( α \) = degree of dissociation

### Step 5: Rearranging the Equation to Find α
From the equation \( i = 1 + (n - 1)α \), we can rearrange it to solve for α:
\[ i - 1 = (n - 1)α \]
\[ α = \frac{i - 1}{n - 1} \]

### Step 6: Final Expression for Degree of Dissociation
Thus, the degree of dissociation (α) of a weak electrolyte can be expressed as:
\[ α = \frac{i - 1}{n - 1} \]

### Final Answer
The degree of dissociation \( α \) of a weak electrolyte is given by:
\[ α = \frac{i - 1}{n - 1} \]