Van't Hoff factor of `Ca(NO_(3))_(2)` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the Van't Hoff factor (i) for the compound \( \text{Ca(NO}_3\text{)}_2 \), we need to analyze how the compound dissociates in solution.

### Step-by-Step Solution:

1. **Identify the Compound**: The compound in question is \( \text{Ca(NO}_3\text{)}_2 \).

2. **Dissociation in Solution**: When \( \text{Ca(NO}_3\text{)}_2 \) dissolves in water, it dissociates into its constituent ions.

The dissociation can be represented as:
\[ \text{Ca(NO}_3\text{)}_2 \rightarrow \text{Ca}^{2+} + 2 \text{NO}_3^{-} \]

3. **Count the Ions Produced**: From the dissociation equation, we can see that:
- 1 calcium ion (\( \text{Ca}^{2+} \))
- 2 nitrate ions (\( \text{NO}_3^{-} \))

Therefore, the total number of ions produced upon dissociation is:
\[ 1 + 2 = 3 \]

4. **Determine the Van't Hoff Factor (i)**: The Van't Hoff factor (i) is defined as the total number of particles (ions) in solution after dissociation. In this case, since we have a total of 3 ions, the Van't Hoff factor is:
\[ i = 3 \]

### Conclusion:
The Van't Hoff factor for \( \text{Ca(NO}_3\text{)}_2 \) is **3**.