`5 cm^(3)` of acetone is added to `100 cm^(3)` of water . Then the vapour pressure of the vapour pressure of the solution _______.

Options

A. It will be equal to the vapour pressure of pure water
B. It will be less than the vapour pressure of pure water
C. It will be greater than the vapour pressure of pure water
D. It will be very large

Answer

To solve the problem of determining the vapor pressure of a solution formed by mixing 5 cm³ of acetone with 100 cm³ of water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- We have two components: acetone (C₃H₆O) and water (H₂O).
- Acetone is the solute, and water is the solvent due to the larger volume of water.

2. **Understand Vapor Pressure**:
- Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature.
- The presence of a solute (acetone) in a solvent (water) typically affects the vapor pressure of the solvent.

3. **Raoult's Law**:
- According to Raoult's Law, the vapor pressure of the solvent in a solution is proportional to the mole fraction of the solvent in the solution.
- The formula is:
\[ P_{solution} = X_{solvent} \cdot P^0_{solvent} \]
where \( P_{solution} \) is the vapor pressure of the solution, \( X_{solvent} \) is the mole fraction of the solvent, and \( P^0_{solvent} \) is the vapor pressure of the pure solvent.

4. **Calculate Mole Fractions**:
- To find the mole fraction of water, we need to calculate the number of moles of water and acetone.
- Assume the density of water is approximately 1 g/cm³, so 100 cm³ of water has a mass of 100 g. The molar mass of water is 18 g/mol, so:
\[ n_{water} = \frac{100 \, g}{18 \, g/mol} \approx 5.56 \, mol \]
- The density of acetone is approximately 0.79 g/cm³, so 5 cm³ of acetone has a mass of:
\[ 5 \, cm³ \times 0.79 \, g/cm³ = 3.95 \, g \]
The molar mass of acetone is 58 g/mol, so:
\[ n_{acetone} = \frac{3.95 \, g}{58 \, g/mol} \approx 0.068 \, mol \]

5. **Calculate Total Moles**:
- Total moles in the solution:
\[ n_{total} = n_{water} + n_{acetone} = 5.56 \, mol + 0.068 \, mol \approx 5.628 \, mol \]

6. **Calculate Mole Fraction of Water**:
- The mole fraction of water \( X_{water} \) is:
\[ X_{water} = \frac{n_{water}}{n_{total}} = \frac{5.56}{5.628} \approx 0.987 \]

7. **Determine Vapor Pressure of the Solution**:
- Since the mole fraction of water is very high, the vapor pressure of the solution will be very close to the vapor pressure of pure water.
- Therefore, the vapor pressure of the solution will be less than the vapor pressure of pure water due to the presence of the solute (acetone).

8. **Conclusion**:
- The vapor pressure of the solution will be less than the vapor pressure of pure water.

### Final Answer:
The vapor pressure of the solution is **less than the vapor pressure of pure water**.