The van't Hoff factor for aqueous solution of `CuSO_(4).5H_(2)O` will be..........

Answer

To determine the van't Hoff factor (i) for the aqueous solution of `CuSO4·5H2O`, we can follow these steps:

### Step 1: Understand the composition of `CuSO4·5H2O`
`CuSO4·5H2O` is copper(II) sulfate pentahydrate. It consists of one formula unit of copper sulfate (CuSO4) and five water molecules (5H2O).

### Step 2: Dissociation of `CuSO4`
When `CuSO4` is dissolved in water, it dissociates into its constituent ions:
- Copper ion (Cu²⁺)
- Sulfate ion (SO₄²⁻)

### Step 3: Count the number of ions produced
From the dissociation of `CuSO4`, we get:
- 1 Cu²⁺ ion
- 1 SO₄²⁻ ion

Thus, the total number of ions produced from the dissociation of `CuSO4` is:
1 (Cu²⁺) + 1 (SO₄²⁻) = 2 ions

### Step 4: Determine the van't Hoff factor (i)
The van't Hoff factor (i) is defined as the total number of particles (ions) into which a solute dissociates in solution. Therefore, for `CuSO4`, the van't Hoff factor is:
i = 2

### Conclusion
The van't Hoff factor for the aqueous solution of `CuSO4·5H2O` is **2**.

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