The van't hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide (Ba(OH)₂), we can follow these steps:

### Step 1: Understand the Dissociation of Barium Hydroxide
Barium hydroxide is a strong electrolyte, which means it completely dissociates in solution. The dissociation can be represented as:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

### Step 2: Determine the Number of Particles Produced
When 1 mole of barium hydroxide dissociates, it produces:
- 1 mole of barium ions (Ba²⁺)
- 2 moles of hydroxide ions (OH⁻)

Thus, from 1 mole of Ba(OH)₂, we get a total of:
\[ 1 \text{ mole (Ba}^{2+}) + 2 \text{ moles (OH}^-) = 3 \text{ moles of ions} \]

### Step 3: Calculate the Van't Hoff Factor (i)
The van't Hoff factor (i) is defined as the total number of particles in solution after dissociation divided by the number of formula units initially dissolved. In this case:
- Initial moles of Ba(OH)₂ = 1 mole
- Total moles at equilibrium = 3 moles (1 Ba²⁺ + 2 OH⁻)

Thus, the van't Hoff factor (i) can be calculated as:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} = \frac{3}{1} = 3 \]

### Conclusion
The van't Hoff factor (i) for a dilute aqueous solution of barium hydroxide is **3**.

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