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 and water mixture as anti-freezing liquid in the radiator of automobiles.
A solution `M` is prepared by mixing ethanol and water. Thus moel 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 `= 155.7K`
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`
Molecualr weight of water `= 18 g mol^(-1)`
Molecular weight of ethanol `= 46 g mol^(-1)`
In asweering 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 the solution M prepared by mixing ethanol and water, we will apply Raoult's Law, which states that the vapor pressure of a solution is equal to the sum of the partial vapor pressures of each component in the solution.

### Step-by-Step Solution:

1. **Identify the Mole Fractions**:
- Given that the mole fraction of ethanol (x_ethanol) is 0.9, we can find the mole fraction of water (x_water):
\[ x_{water} = 1 - x_{ethanol} = 1 - 0.9 = 0.1 \]

2. **Determine the Vapor Pressures of Pure Solvents**:
- The vapor pressure of pure ethanol (P°_ethanol) is given as 40 mm Hg.
- The vapor pressure of pure water (P°_water) is given as 32.8 mm Hg.

3. **Calculate the Partial Vapor Pressures**:
- Using Raoult's Law, the partial vapor pressure of ethanol (P_ethanol) in the solution is calculated as:
\[ P_{ethanol} = P°_{ethanol} \times x_{ethanol} = 40 \, \text{mm Hg} \times 0.9 = 36 \, \text{mm Hg} \]
- The partial vapor pressure of water (P_water) in the solution is:
\[ P_{water} = P°_{water} \times x_{water} = 32.8 \, \text{mm Hg} \times 0.1 = 3.28 \, \text{mm Hg} \]

4. **Calculate the Total Vapor Pressure of the Solution**:
- The total vapor pressure (P_total) of the solution M is the sum of the partial vapor pressures:
\[ P_{total} = P_{ethanol} + P_{water} = 36 \, \text{mm Hg} + 3.28 \, \text{mm Hg} = 39.28 \, \text{mm Hg} \]

### Final Answer:
The vapor pressure of the solution M is approximately **39.28 mm Hg**.