Consider following solution:
0.1 m`C_(6)H_(5)NH_(3^(+))Cl^(-)`, 0.1 m Kcl, 0.1 m Glucose, 0.1 m `Na_(2)C_(2)O_(4).10H_(2)O`

Options

A. the solution with higher boiling point is 0.1 `Na_(2)C_(2)O_(4).10H_(2)O`
B. the solution with higher freezing point is 0.1 m glucose
C. 0.1 m `C_(6)H_(5)NH_(3)Cl` and 0.1 m NaClwill have the same osmotic pressure
D. 0.1 m glucose solution will have the lowest osmotic pressure

Answer

To solve the question regarding the given solutions, we will analyze each solution based on the Van't Hoff factor (I), which determines the colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure.

### Step-by-Step Solution:

1. **Identify the Solutions and Their Concentrations:**
- 0.1 m `C6H5NH3+` Cl⁻
- 0.1 m KCl
- 0.1 m Glucose
- 0.1 m `Na2C2O4.10H2O`

2. **Calculate the Van't Hoff Factor (I) for Each Solution:**
- **For `C6H5NH3+` Cl⁻:**
- This dissociates into 2 ions: `C6H5NH3+` and Cl⁻.
- Therefore, I = 2.

- **For KCl:**
- This dissociates into 2 ions: K⁺ and Cl⁻.
- Therefore, I = 2.

- **For Glucose:**
- Glucose does not dissociate in solution; it remains as one molecule.
- Therefore, I = 1.

- **For `Na2C2O4.10H2O`:**
- This dissociates into 3 ions: 2 Na⁺ and C2O4²⁻.
- Therefore, I = 3.

3. **Determine the Elevation in Boiling Point (ΔTb):**
- The formula for boiling point elevation is ΔTb = Kb × m × I.
- Since all solutions have the same molality (0.1 m), the elevation in boiling point will depend solely on the Van't Hoff factor (I).
- The solution with the highest I will have the highest boiling point.
- From our calculations:
- `C6H5NH3+` Cl⁻: I = 2
- KCl: I = 2
- Glucose: I = 1
- `Na2C2O4.10H2O`: I = 3
- Therefore, `Na2C2O4.10H2O` has the highest boiling point.

4. **Determine the Depression in Freezing Point (ΔTf):**
- The formula for freezing point depression is ΔTf = Kf × m × I.
- Similar to boiling point elevation, the freezing point depression depends on the Van't Hoff factor (I).
- The solution with the lowest I will have the highest freezing point.
- Glucose has the lowest I = 1, hence it will have the highest freezing point.

5. **Determine the Osmotic Pressure (π):**
- The formula for osmotic pressure is π = C × R × T × I.
- Since C, R, and T are constants, the osmotic pressure will depend on I.
- Comparing the I values:
- `C6H5NH3+` Cl⁻: I = 2
- KCl: I = 2
- Glucose: I = 1
- `Na2C2O4.10H2O`: I = 3
- Therefore, `C6H5NH3+` Cl⁻ and KCl will have the same osmotic pressure as they have the same I.

6. **Conclusion:**
- The solution with the highest boiling point is `Na2C2O4.10H2O`.
- The solution with the highest freezing point is glucose.
- `C6H5NH3+` Cl⁻ and KCl have the same osmotic pressure.
- Glucose has the lowest osmotic pressure.

### Final Answer:
All statements regarding the solutions are correct:
- A: The solution with the higher boiling point is `Na2C2O4.10H2O`.
- B: The solution with the higher freezing point is glucose.
- C: `C6H5NH3+` Cl⁻ and KCl will have the same osmotic pressure.
- D: The glucose solution will have the lowest osmotic pressure.