Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. one of its examples is the use of ethylene glycol adn water mixtures as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is `0.9`.
Given: Freezing point depression constant of water `(K_(f)^(water)) = 1.86 K kg mol^(-1)`
Freezing point depression constant of ethanol `(K_(f)^("ethanol")) = 2.0 K kg mol^(-1)`
Boiling point elevation constant of water `(K_(b)^(water)) = 0.52 K kg mol^(-1)`
Boiling point elevation constant of ethanol `(K_(b)^("ethanol")) = 1.2 K kg mol^(-1)`
Standard freezing point of water `= 273 K`
Standard freezing point of ethanol `= 155.7 K`
Standard boiling point of water `= 373 K`
Standard boiling point of ethanol `= 351.5 K`
Vapour pressure of pure water `= 32.8 mm Hg`
Vapour pressure of pure ethanol `= 40 mm Hg`
Molecular weight of water ` =18 g mol^(-1)`
Molecular weight of ethanol `= 46 g mol^(-1)`
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
The vapour pressure of the solution M is

Options

A. `39.3 mm Hg`
B. `36.0 mm Hg`
C. `29.5 mm Hg`
D. `28.8 mm Hg`

Explanation

To find the vapor pressure of solution M prepared by mixing ethanol and water with a mole fraction of ethanol of 0.9, we will use Raoult's Law. Raoult's Law states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

### Step-by-step Solution:

1. **Identify the Components**:
- The solution is made of ethanol (C₂H₅OH) and water (H₂O).
- The mole fraction of ethanol (X_ethanol) = 0.9.
- Therefore, the mole fraction of water (X_water) = 1 - X_ethanol = 1 - 0.9 = 0.1.

2. **Vapor Pressure of Pure Solvent**:
- The vapor pressure of pure ethanol (P°_ethanol) = 40 mm Hg.
- Since ethanol has a higher mole fraction in the solution, we will consider it as the solvent for our calculations.

3. **Apply Raoult's Law**:
- According to Raoult's Law:
\[ P_{solution} = P°_{ethanol} \times X_{ethanol} \]
- Substitute the values:
\[ P_{solution} = 40 \, \text{mm Hg} \times 0.9 \]

4. **Calculate the Vapor Pressure**:
- Perform the multiplication:
\[ P_{solution} = 40 \times 0.9 = 36 \, \text{mm Hg} \]

5. **Final Answer**:
- The vapor pressure of solution M is **36 mm Hg**.

### Summary of Calculation:
- Vapor pressure of solution M = 36 mm Hg.