The degree of dissociation `(alpha)` of a weak electrolyte, `A_(2)SO_(4)` is related to Vant Hoff factor (i) by the expression

Options

A. `i=1+2prop`
B. `i=1+3prop`
C. `i=1+prop`
D. `i=1+4prop`

Answer

To solve the problem of finding the relationship between the degree of dissociation (α) of the weak electrolyte \( A_2SO_4 \) and the Van't Hoff factor (i), we can follow these steps:

### Step 1: Understand the dissociation of the electrolyte
The weak electrolyte \( A_2SO_4 \) dissociates in solution as follows:
\[ A_2SO_4 \rightleftharpoons 2A^+ + SO_4^{2-} \]
Here, \( A_2SO_4 \) dissociates into 2 ions of \( A^+ \) and 1 ion of \( SO_4^{2-} \).

### Step 2: Define the degree of dissociation (α)
Let \( \alpha \) be the degree of dissociation. If we start with 1 mole of \( A_2SO_4 \):
- The amount of \( A_2SO_4 \) that remains after dissociation is \( 1 - \alpha \).
- The amount of \( A^+ \) produced is \( 2\alpha \).
- The amount of \( SO_4^{2-} \) produced is \( \alpha \).

### Step 3: Calculate the total number of particles after dissociation
The total number of particles after dissociation can be calculated as:
\[ \text{Total particles} = (1 - \alpha) + 2\alpha + \alpha = 1 + 2\alpha \]

### Step 4: Define the Van't Hoff factor (i)
The Van't Hoff factor (i) is defined as the ratio of the total number of particles in solution to the number of formula units initially dissolved:
\[ i = \frac{\text{Total particles}}{\text{Initial particles}} = \frac{1 + 2\alpha}{1} \]
Thus, we have:
\[ i = 1 + 2\alpha \]

### Step 5: Rearranging the equation to find α in terms of i
From the equation \( i = 1 + 2\alpha \), we can rearrange it to find α:
\[ 2\alpha = i - 1 \]
\[ \alpha = \frac{i - 1}{2} \]

### Final Expression
The relationship between the degree of dissociation (α) and the Van't Hoff factor (i) for the weak electrolyte \( A_2SO_4 \) is:
\[ \alpha = \frac{i - 1}{2} \]