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

### Step 1: Understand the Concept of Vapor Pressure Lowering
The lowering of vapor pressure is a colligative property, which means it depends on the number of solute particles in the solution rather than the identity of the solute. The formula for the relative lowering of vapor pressure is given by:

\[ \frac{\Delta P}{P_0} = i \cdot m \]

where:
- \(\Delta P\) is the change in vapor pressure,
- \(P_0\) is the vapor pressure of the pure solvent,
- \(i\) is the Van't Hoff factor (number of particles the solute dissociates into),
- \(m\) is the molality of the solution.

### Step 2: Identify the Van't Hoff Factor for Each Solute
1. **For NaCl**:
- NaCl dissociates into Na⁺ and Cl⁻.
- Therefore, \(i = 2\).

2. **For CuSO4**:
- CuSO4 dissociates into Cu²⁺ and SO₄²⁻.
- Therefore, \(i = 2\).

3. **For K2SO4**:
- K2SO4 dissociates into 2 K⁺ and SO₄²⁻.
- Therefore, \(i = 3\).

### Step 3: Calculate the Relative Lowering of Vapor Pressure
Since the molality (m) is constant at 0.1 M for all solutions, we can express the relative lowering of vapor pressure for each solute.

- For NaCl:
\[ \frac{\Delta P}{P_0} = 2 \cdot 0.1 = 0.2 \]

- For CuSO4:
\[ \frac{\Delta P}{P_0} = 2 \cdot 0.1 = 0.2 \]

- For K2SO4:
\[ \frac{\Delta P}{P_0} = 3 \cdot 0.1 = 0.3 \]

### Step 4: Compare the Lowering of Vapor Pressure
Now we can compare the lowering of vapor pressure for the three solutions:
- NaCl: 0.2
- CuSO4: 0.2
- K2SO4: 0.3

### Step 5: Express the Ratios
The ratios of the lowering of vapor pressure can be expressed as:
- NaCl : CuSO4 : K2SO4 = 0.2 : 0.2 : 0.3

To simplify this ratio:
- Divide each term by 0.1:
\[ 2 : 2 : 3 \]
- This can be further simplified to:
\[ 1 : 1 : 1.5 \]

### Conclusion
Thus, the lowering of vapor pressure for the given solutions is in the ratio of 1 : 1 : 1.5. The correct answer corresponds to option 2.

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