The Van't Hoff factor of a `0.1 M Al_(2)(SO_(4))_(3)` solution is `4.20`. The degree of dissociation is

Options

A. `80%`
B. `90%`
C. `78%`
D. `83%`

Explanation

To find the degree of dissociation (α) of a 0.1 M solution of \( Al_2(SO_4)_3 \) given that the Van't Hoff factor (i) is 4.20, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissociation of \( Al_2(SO_4)_3 \)**:
The dissociation of aluminum sulfate can be represented as:
\[ Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \]
From this equation, we can see that 1 mole of \( Al_2(SO_4)_3 \) produces 2 moles of \( Al^{3+} \) ions and 3 moles of \( SO_4^{2-} \) ions, totaling 5 moles of ions.

2. **Setting Up the Expression for Van't Hoff Factor (i)**:
The Van't Hoff factor (i) is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved. If α is the degree of dissociation, then:
\[ i = 1 - \alpha + 2\alpha + 3\alpha = 1 + 4\alpha \]
Here, \( 1 - \alpha \) is the undissociated \( Al_2(SO_4)_3 \), \( 2\alpha \) is the dissociated \( Al^{3+} \) ions, and \( 3\alpha \) is the dissociated \( SO_4^{2-} \) ions.

3. **Substituting the Given Value of i**:
We know from the problem statement that \( i = 4.20 \). Therefore, we can set up the equation:
\[ 4.20 = 1 + 4\alpha \]

4. **Solving for α**:
Rearranging the equation gives:
\[ 4\alpha = 4.20 - 1 \]
\[ 4\alpha = 3.20 \]
\[ \alpha = \frac{3.20}{4} = 0.80 \]

5. **Converting α to Percentage**:
To express the degree of dissociation as a percentage, we multiply by 100:
\[ \text{Degree of dissociation} = \alpha \times 100 = 0.80 \times 100 = 80\% \]

### Final Answer:
The degree of dissociation (α) of the \( 0.1 M \) \( Al_2(SO_4)_3 \) solution is **80%**.