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 solve the problem regarding the lowering of vapor pressure of 0.1 M aqueous solutions of NaCl, CuSO₄, and K₂SO₄, we will follow these steps:

### Step 1: Understand the Concept of Lowering of Vapor Pressure
The lowering of vapor pressure in a solution is directly proportional to the number of solute particles in the solution. This relationship is described by Raoult's Law, which states that the vapor pressure of a solvent is lowered when a non-volatile solute is added.

### Step 2: Identify the Van't Hoff Factor (i)
The Van't Hoff factor (i) is defined as the number of particles into which a solute dissociates in solution. It is crucial for calculating the lowering of vapor pressure.

### Step 3: Calculate the Van't Hoff Factor for Each Solute
1. **For NaCl**:
- NaCl dissociates into Na⁺ and Cl⁻.
- Number of moles before dissociation = 1 (NaCl).
- Number of moles after dissociation = 2 (Na⁺ + Cl⁻).
- Van't Hoff factor (i) for NaCl = 2.

2. **For CuSO₄**:
- CuSO₄ dissociates into Cu²⁺ and SO₄²⁻.
- Number of moles before dissociation = 1 (CuSO₄).
- Number of moles after dissociation = 2 (Cu²⁺ + SO₄²⁻).
- Van't Hoff factor (i) for CuSO₄ = 2.

3. **For K₂SO₄**:
- K₂SO₄ dissociates into 2K⁺ and SO₄²⁻.
- Number of moles before dissociation = 1 (K₂SO₄).
- Number of moles after dissociation = 3 (2K⁺ + SO₄²⁻).
- Van't Hoff factor (i) for K₂SO₄ = 3.

### Step 4: Determine the Ratio of Lowering of Vapor Pressure
Since the lowering of vapor pressure is directly proportional to the Van't Hoff factor, we can express the ratio of lowering of vapor pressure for the three solutions as follows:
- For NaCl: i = 2
- For CuSO₄: i = 2
- For K₂SO₄: i = 3

Thus, the ratio of lowering of vapor pressure can be expressed as:
\[ \text{Ratio} = \text{NaCl} : \text{CuSO₄} : \text{K₂SO₄} = 2 : 2 : 3 \]

### Step 5: Simplify the Ratio
To express the ratio in simpler terms:
- Divide all parts of the ratio by 2:
\[ \text{Ratio} = 1 : 1 : 1.5 \]

### Conclusion
The final ratio of lowering of vapor pressure for the 0.1 M aqueous solutions of NaCl, CuSO₄, and K₂SO₄ is:
\[ \text{NaCl} : \text{CuSO₄} : \text{K₂SO₄} = 1 : 1 : 1.5 \]