The lowering of vapour pressure of 0.1 M aqueous solution `NaCl, CuSO_(4) and K_(2)SO_(4)` are :

Options

A. All equal
B. In the ratio of `1 : 1: 15`
C. In the ratio 3 : 2 : 1
D. In the ratio of 1.5 : 1 : 2.5

Answer

To find the lowering of vapor pressure of 0.1 M aqueous solutions of NaCl, CuSO₄, and K₂SO₄, we will use the concept of colligative properties, specifically focusing on the Van't Hoff factor (i) for each solute. The lowering of vapor pressure is directly related to the number of particles that a solute dissociates into when dissolved in a solvent.

### Step-by-Step Solution:

1. **Understand Colligative Properties**:
- Colligative properties depend on the number of solute particles in a solution, not on the identity of the solute. The lowering of vapor pressure is one such property.

2. **Identify the Van't Hoff Factor (i)**:
- The Van't Hoff factor (i) indicates the number of particles into which a solute dissociates in solution.
- For each solute:
- **NaCl**: Dissociates into Na⁺ and Cl⁻. Thus, i = 2 (1 Na⁺ + 1 Cl⁻).
- **CuSO₄**: Dissociates into Cu²⁺ and SO₄²⁻. Thus, i = 2 (1 Cu²⁺ + 1 SO₄²⁻).
- **K₂SO₄**: Dissociates into 2 K⁺ and SO₄²⁻. Thus, i = 3 (2 K⁺ + 1 SO₄²⁻).

3. **Calculate the Lowering of Vapor Pressure**:
- The lowering of vapor pressure (ΔP) can be expressed as:
\[ \Delta P = i \cdot K \cdot C \]
where K is a constant and C is the molarity of the solution.
- Since all solutions are 0.1 M, we can focus on the Van't Hoff factor (i) for comparison.

4. **List the Van't Hoff Factors**:
- NaCl: i = 2
- CuSO₄: i = 2
- K₂SO₄: i = 3

5. **Compare the Lowering of Vapor Pressure**:
- Since the lowering of vapor pressure is proportional to the Van't Hoff factor:
- NaCl: ΔP ∝ 2
- CuSO₄: ΔP ∝ 2
- K₂SO₄: ΔP ∝ 3

6. **Conclusion**:
- The order of lowering of vapor pressure for the given solutions is:
- K₂SO₄ > NaCl = CuSO₄

### Final Answer:
- The lowering of vapor pressure for 0.1 M aqueous solutions of NaCl, CuSO₄, and K₂SO₄ is:
- NaCl: 2
- CuSO₄: 2
- K₂SO₄: 3