For the given electrolyte `A_(x)B_(y)`. The degree of dissociation `'alpha'` can be given as:

Options

A. `alpha=(i-1)/(x+y-1)`
B. `i=(1-alpha)+xalpha+yalpha`
C. `alpha=(1-i)/(1-x-y)`
D. none

Explanation

To find the degree of dissociation (α) for the electrolyte \( A_xB_y \), we can follow these steps:

### Step 1: Understand the dissociation of the electrolyte
The electrolyte \( A_xB_y \) dissociates in solution to form \( x \) moles of \( A \) and \( y \) moles of \( B \). The dissociation can be represented as:
\[ A_xB_y \rightleftharpoons xA + yB \]

### Step 2: Define the degree of dissociation
Let \( \alpha \) be the degree of dissociation. This means that if we start with 1 mole of \( A_xB_y \), after dissociation, \( \alpha \) moles of \( A_xB_y \) will dissociate.

### Step 3: Calculate the moles after dissociation
Initially, we have:
- 1 mole of \( A_xB_y \)
- After dissociation, the moles of \( A \) formed will be \( x\alpha \) and the moles of \( B \) formed will be \( y\alpha \).

Thus, the total number of moles after dissociation can be expressed as:
\[ \text{Total moles} = 1 - \alpha + x\alpha + y\alpha = 1 + (x + y - 1)\alpha \]

### Step 4: Calculate the van 't Hoff factor (i)
The van 't Hoff factor \( i \) is defined as the total number of particles in solution after dissociation. Therefore, we have:
\[ i = \text{Total moles after dissociation} = 1 + (x + y - 1)\alpha \]

### Step 5: Relate degree of dissociation to van 't Hoff factor
From the expression for \( i \):
\[ i = 1 + (x + y - 1)\alpha \]
Rearranging gives:
\[ \alpha = \frac{i - 1}{x + y - 1} \]

### Step 6: Alternative expression for degree of dissociation
We can also express \( i \) in terms of \( \alpha \):
\[ i = 1 - \alpha + x\alpha + y\alpha = 1 + (x + y - 1)\alpha \]

### Final Expression for Degree of Dissociation
Thus, we can conclude that the degree of dissociation \( \alpha \) can be expressed as:
\[ \alpha = \frac{i - 1}{x + y - 1} \]
or alternatively,
\[ \alpha = \frac{1 - i}{1 - (x + y)} \]

### Conclusion
The degree of dissociation \( \alpha \) for the electrolyte \( A_xB_y \) can be expressed using the above equations, indicating that all forms provided in the options can be correct depending on the context.