To solve the problem of finding the vapor pressure of solution M, which is prepared by mixing ethanol and water, we can follow these steps:
### Step-by-Step Solution:
1. **Identify the Mole Fractions**:
- Given that the mole fraction of ethanol (C2H5OH) is \( x_{ethanol} = 0.9 \).
- Therefore, the mole fraction of water (H2O) is:
\[
x_{water} = 1 - x_{ethanol} = 1 - 0.9 = 0.1
\]
2. **Understand the Vapor Pressure Contribution**:
- The vapor pressure of the solution can be calculated using Raoult's Law, which states:
\[
P_{solution} = P_{ethanol}^{\circ} \cdot x_{ethanol} + P_{water}^{\circ} \cdot x_{water}
\]
- Where:
- \( P_{ethanol}^{\circ} \) is the vapor pressure of pure ethanol.
- \( P_{water}^{\circ} \) is the vapor pressure of pure water.
3. **Substitute the Known Values**:
- From the problem, we have:
- \( P_{ethanol}^{\circ} = 40 \, \text{mmHg} \)
- \( P_{water}^{\circ} = 32.8 \, \text{mmHg} \)
- Now substitute the values into the equation:
\[
P_{solution} = (40 \, \text{mmHg} \cdot 0.9) + (32.8 \, \text{mmHg} \cdot 0.1)
\]
4. **Calculate Each Term**:
- Calculate the contribution from ethanol:
\[
40 \, \text{mmHg} \cdot 0.9 = 36 \, \text{mmHg}
\]
- Calculate the contribution from water:
\[
32.8 \, \text{mmHg} \cdot 0.1 = 3.28 \, \text{mmHg}
\]
5. **Add the Contributions**:
- Now, add the contributions from both components:
\[
P_{solution} = 36 \, \text{mmHg} + 3.28 \, \text{mmHg} = 39.28 \, \text{mmHg}
\]
6. **Final Result**:
- Since the problem states that we are considering ideal dilute solutions and that the solute (water) is non-volatile, we can ignore the contribution from water. Therefore, the vapor pressure of the solution is primarily due to ethanol:
\[
P_{solution} \approx 36 \, \text{mmHg}
\]
### Conclusion:
The vapor pressure of the solution M is approximately **36 mmHg**.
---
