In which of the following the vant Hoff factor (i) is equal to one:

Options

A. `NaCI`
B. `KNO_(3)`
C. Urea
D. all

Answer

To determine which of the given compounds has a van't Hoff factor (i) equal to one, we need to analyze the dissociation of each compound in solution. The van't Hoff factor (i) represents the number of particles the solute breaks into when dissolved in a solvent.

Let's evaluate each option step by step:

### Step 1: Analyze NaCl
- **Dissociation**: NaCl dissociates in water into Na⁺ and Cl⁻ ions.
- **Calculation**:
- NaCl → Na⁺ + Cl⁻
- This results in 2 particles.
- **van't Hoff factor (i)**: i = 2

### Step 2: Analyze KNO3
- **Dissociation**: KNO3 dissociates in water into K⁺ and NO3⁻ ions.
- **Calculation**:
- KNO3 → K⁺ + NO3⁻
- This also results in 2 particles.
- **van't Hoff factor (i)**: i = 2

### Step 3: Analyze Urea
- **Dissociation**: Urea (NH2CONH2) is a covalent compound and does not dissociate into ions in solution.
- **Calculation**:
- Urea does not break into smaller particles.
- **van't Hoff factor (i)**: i = 1

### Step 4: Analyze Glucose
- **Dissociation**: Glucose (C6H12O6) is also a covalent compound and does not dissociate into ions in solution.
- **Calculation**:
- Glucose does not break into smaller particles.
- **van't Hoff factor (i)**: i = 1

### Conclusion:
From the analysis, we find that:
- NaCl has i = 2
- KNO3 has i = 2
- Urea has i = 1
- Glucose has i = 1

Thus, the compounds with a van't Hoff factor equal to one are **Urea and Glucose**.

### Final Answer:
Urea and Glucose have a van't Hoff factor (i) equal to 1.