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 reaction can be written as:
\[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{OH}^- \]

### Step 2: Count the Number of Ions Produced
From the dissociation equation, we can see that one formula unit of barium hydroxide produces:
- 1 barium ion (\(\text{Ba}^{2+}\))
- 2 hydroxide ions (\(\text{OH}^-\))

Thus, the total number of ions produced from one unit of barium hydroxide is:
\[ n = 1 + 2 = 3 \]

### Step 3: Use the Formula for van't Hoff Factor
The van't Hoff factor (i) can be defined in terms of the degree of dissociation (α) and the number of ions (n) produced:
\[ \alpha = \frac{i - 1}{n - 1} \]

For a strong electrolyte like barium hydroxide, the degree of dissociation (α) is equal to 1 because it completely dissociates. Therefore, we can substitute α = 1 into the equation:
\[ 1 = \frac{i - 1}{n - 1} \]

### Step 4: Substitute the Value of n
We have already determined that \( n = 3 \). Substituting this value into the equation gives:
\[ 1 = \frac{i - 1}{3 - 1} \]
\[ 1 = \frac{i - 1}{2} \]

### Step 5: Solve for i
Now, we can solve for i:
\[ 2 = i - 1 \]
\[ i = 2 + 1 \]
\[ i = 3 \]

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