An electrolyte `A_(2)B_(3)` ionizes in water upto 75%. The van't Hoff factor for it is .

Explanation

To solve the problem, we need to determine the van't Hoff factor (i) for the electrolyte A₂B₃ that ionizes in water up to 75%. Let's go through the solution step by step:

### Step 1: Understand the dissociation of the electrolyte
The electrolyte A₂B₃ dissociates in water into its ions. The dissociation can be represented as:
\[ \text{A}_2\text{B}_3 \rightarrow 2\text{A}^{3+} + 3\text{B}^{2-} \]
This means that one formula unit of A₂B₃ produces a total of 5 ions (2 A³⁺ ions and 3 B²⁻ ions).

### Step 2: Define the ionization degree (α)
The problem states that the electrolyte ionizes up to 75%. This means:
\[ \alpha = 0.75 \]
where α is the fraction of the electrolyte that has ionized.

### Step 3: Calculate the number of particles before and after dissociation
- **Initial number of particles (ni)**: Before dissociation, we have 1 mole of A₂B₃, so:
\[ n_i = 1 \]

- **Final number of particles (nf)**: After dissociation, we need to calculate how many particles are present. The amount of A₂B₃ that dissociates is α moles, which means:
- Remaining A₂B₃ = \( 1 - \alpha = 1 - 0.75 = 0.25 \) moles
- A³⁺ ions produced = \( 2\alpha = 2 \times 0.75 = 1.5 \) moles
- B²⁻ ions produced = \( 3\alpha = 3 \times 0.75 = 2.25 \) moles

Thus, the total number of particles after dissociation is:
\[ n_f = (1 - \alpha) + 2\alpha + 3\alpha = (1 - 0.75) + 2(0.75) + 3(0.75) \]
\[ n_f = 0.25 + 1.5 + 2.25 = 4 \]

### Step 4: Calculate the van't Hoff factor (i)
The van't Hoff factor (i) is defined as:
\[ i = \frac{n_f}{n_i} \]
Substituting the values we calculated:
\[ i = \frac{4}{1} = 4 \]

### Final Answer
Thus, the van't Hoff factor for the electrolyte A₂B₃ that ionizes in water up to 75% is:
\[ \boxed{4} \]

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