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. Application 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. 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 `(e_(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 pressusre 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, which is a mixture of ethanol and water, we can use Raoult's Law. According to Raoult's Law, the total vapor pressure of a solution (P_total) is the sum of the partial pressures of each component in the solution, which can be calculated using their mole fractions and their respective vapor pressures in pure states.

### Step-by-Step Solution:

1. **Identify the Mole Fractions:**
- Given 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. **Identify the Vapor Pressures of Pure Components:**
- The vapor pressure of pure ethanol (P°_ethanol) = 40 mm Hg.
- The vapor pressure of pure water (P°_water) = 32.8 mm Hg.

3. **Calculate the Partial Pressure of Ethanol:**
- Using Raoult's Law, the partial pressure of ethanol (P_ethanol) can be calculated as:
\[ P_{\text{ethanol}} = X_{\text{ethanol}} \times P°_{\text{ethanol}} = 0.9 \times 40 \, \text{mm Hg} = 36 \, \text{mm Hg} \]

4. **Calculate the Partial Pressure of Water:**
- The partial pressure of water (P_water) can be calculated as:
\[ P_{\text{water}} = X_{\text{water}} \times P°_{\text{water}} = 0.1 \times 32.8 \, \text{mm Hg} = 3.28 \, \text{mm Hg} \]

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

### Final Answer:
The vapor pressure of solution M is approximately **39.3 mm Hg**.