IF a=50% for `Al_2(SO_4)_3` then van't hoff factor would be

Options

A. 1.5
B. 0.5
C. 3
D. 2

Answer

To find the van 't Hoff factor (i) for \( Al_2(SO_4)_3 \) when the degree of dissociation (α) is 50%, we can follow these steps:

### Step 1: Understand the dissociation of \( Al_2(SO_4)_3 \)
The compound \( Al_2(SO_4)_3 \) dissociates in solution as follows:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]
This means that one formula unit of \( Al_2(SO_4)_3 \) produces a total of 5 ions (2 aluminum ions and 3 sulfate ions).

### Step 2: Determine the number of ions (n)
From the dissociation, we can see that:
- \( n = 5 \) (the total number of ions produced from one formula unit).

### Step 3: Use the formula for van 't Hoff factor (i)
The van 't Hoff factor (i) is given by the formula:
\[ i = 1 + (n - 1) \cdot \alpha \]
Where:
- \( n \) is the number of ions produced,
- \( \alpha \) is the degree of dissociation.

### Step 4: Substitute the values into the formula
Given that \( \alpha = 0.5 \) (since 50% dissociation) and \( n = 5 \):
\[ i = 1 + (5 - 1) \cdot 0.5 \]
\[ i = 1 + 4 \cdot 0.5 \]
\[ i = 1 + 2 \]
\[ i = 3 \]

### Conclusion
The van 't Hoff factor (i) for \( Al_2(SO_4)_3 \) when the degree of dissociation is 50% is **3**.

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