Which of the following compounds corresponds to van’t Hoff factor (i) to be equal to 2 for its dilute solution?

Options

A. `K_(2)SO_(4)`
B. `NaHSO_(4)`
C. Sugar
D. `MgSO_(4)`

Answer

To determine which compound corresponds to a van’t Hoff factor (i) equal to 2 for its dilute solution, we will analyze the dissociation of each compound provided in the question. The van’t Hoff factor (i) is defined as the total number of particles in solution after dissociation divided by the total number of particles initially present.

### Step-by-Step Solution:

1. **Understanding Van't Hoff Factor (i)**:
The van't Hoff factor (i) is calculated using the formula:
\[ i = \frac{\text{Total number of particles after dissociation}}{\text{Total number of particles initially}} \]
We need to find a compound that gives us \( i = 2 \).

2. **Analyzing Each Compound**:
We will analyze the dissociation of each compound mentioned.

- **K2SO4**:
- Dissociation:
\[ K2SO4 \rightarrow 2K^+ + SO4^{2-} \]
- Initially, there is 1 particle (K2SO4).
- After dissociation, there are 3 particles (2 K\(^+\) and 1 SO4\(^{2-}\)).
- Van't Hoff factor:
\[ i = \frac{3}{1} = 3 \]

- **NaHSO4**:
- Dissociation:
\[ NaHSO4 \rightarrow Na^+ + H^+ + SO4^{2-} \]
- Initially, there is 1 particle (NaHSO4).
- After dissociation, there are 3 particles (1 Na\(^+\), 1 H\(^+\), and 1 SO4\(^{2-}\)).
- Van't Hoff factor:
\[ i = \frac{3}{1} = 3 \]

- **Sugar (C12H22O11)**:
- Sugar does not dissociate in solution.
- Initially, there is 1 particle (C12H22O11).
- After dissociation, there is still 1 particle.
- Van't Hoff factor:
\[ i = \frac{1}{1} = 1 \]

- **MgSO4**:
- Dissociation:
\[ MgSO4 \rightarrow Mg^{2+} + SO4^{2-} \]
- Initially, there is 1 particle (MgSO4).
- After dissociation, there are 2 particles (1 Mg\(^{2+}\) and 1 SO4\(^{2-}\)).
- Van't Hoff factor:
\[ i = \frac{2}{1} = 2 \]

3. **Conclusion**:
The compound that corresponds to a van’t Hoff factor (i) equal to 2 is **MgSO4**.

### Final Answer:
The correct answer is **MgSO4** (D part).