What is the maximum value of van't Hoff factor for` AlCl_(3)` ?

Answer

To find the maximum value of the van't Hoff factor (i) for AlCl₃, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dissociation of AlCl₃**:
AlCl₃ dissociates in water to form aluminum ions and chloride ions. The dissociation can be represented as:
\[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]
From this equation, we can see that one formula unit of AlCl₃ produces one aluminum ion and three chloride ions.

2. **Count the Total Number of Ions Produced**:
When AlCl₃ dissociates completely, it produces:
- 1 Al³⁺ ion
- 3 Cl⁻ ions
Therefore, the total number of ions (n) produced is:
\[ n = 1 + 3 = 4 \]

3. **Understanding the van't Hoff Factor (i)**:
The van't Hoff factor (i) is defined as the number of particles into which a solute dissociates in solution. The formula for calculating the van't Hoff factor is:
\[ i = 1 + n - \alpha \]
where:
- \( n \) is the number of particles produced upon dissociation,
- \( \alpha \) is the degree of dissociation.

4. **Assume Complete Dissociation**:
For the maximum value of the van't Hoff factor, we assume that the dissociation is 100%. This means:
\[ \alpha = 1 \]

5. **Substitute Values into the Formula**:
Now, substituting the values into the van't Hoff factor formula:
\[ i = 1 + n - \alpha \]
Substituting \( n = 4 \) and \( \alpha = 1 \):
\[ i = 1 + 4 - 1 \]
Simplifying this gives:
\[ i = 4 \]

6. **Conclusion**:
The maximum value of the van't Hoff factor for AlCl₃ is 4.

### Final Answer:
The maximum value of the van't Hoff factor for AlCl₃ is **4**.

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