The degree of an electrolyte is a and itsvan't Hoff factor is i. The number of ions obtained by complate dissocition of 1 molecules of electrolyte as :

Options

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

Answer

To solve the problem, we need to find the number of ions obtained from the complete dissociation of one molecule of an electrolyte, given its degree of dissociation (α) and Van't Hoff factor (i).

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The degree of dissociation (α) is the fraction of the total number of electrolyte molecules that dissociate into ions.
- The Van't Hoff factor (i) indicates the number of particles (ions) produced in solution per formula unit of solute.

2. **Establish the Relationship**:
- The relationship between the Van't Hoff factor (i), degree of dissociation (α), and the number of ions (n) can be expressed as:
\[ i = 1 + (n - 1) \cdot \alpha \]
- Here, \(n\) is the number of ions produced from one molecule of the electrolyte.

3. **Rearrange the Equation**:
- Rearranging the equation to solve for \(n\):
\[ i - 1 = (n - 1) \cdot \alpha \]
- This can be rewritten as:
\[ n - 1 = \frac{i - 1}{\alpha} \]
- Adding 1 to both sides gives:
\[ n = \frac{i - 1}{\alpha} + 1 \]

4. **Simplify the Expression**:
- Combine the terms to express \(n\) in a single fraction:
\[ n = \frac{i - 1 + \alpha}{\alpha} \]

5. **Final Expression**:
- Thus, the number of ions obtained by the complete dissociation of one molecule of the electrolyte is:
\[ n = \frac{i + \alpha - 1}{\alpha} \]

### Conclusion:
The final answer to the question is:
\[ n = \frac{i + \alpha - 1}{\alpha} \]