A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given
`k_(f)("water")=1.86Kkg"mol"^(-1)`
`k_(f)("ethanol")=2.0Kkg"mol"^(-1)`
`k_(b)("water")=0.52Kkg"mol"^(-1)`
`k_(b)("ethanol")=1.2Kkg"mol"^(-1)`
Standard f.p. of water = 273 K
Standard f.p. of ethanol = 155.7 K
Standard b.p. of water = 373 K
Standard b.p. of ethanol = 351.5 K
Vapour pressure of pure water = 32.8 mmHg
Vapour pressure of pure ethanol = 40 mmHg
Molecular weight of water = 18 g `"mol"^(-1)`
Molecular weight of ethanol = 46 g `"mol"^(-1)`
consider the solutions to be ideal dilute solutions and solutes to be nonvolatile and non-dissociative.
The vapour pressure of the solution M is

Options

A. 39.3 mmHg
B. 36.0 mmHg
C. 29.5 mmHg
D. 28.8 mmHg

Answer

To find the vapor pressure of the solution M prepared by mixing ethanol and water, we can follow these steps:

### Step 1: Identify the mole fractions
Given that the mole fraction of ethanol (X_ethanol) in the mixture is 0.9, we can calculate the mole fraction of water (X_water):
\[ X_{water} = 1 - X_{ethanol} = 1 - 0.9 = 0.1 \]

### Step 2: Use Raoult's Law
Raoult's Law states that the vapor pressure of a solution (P_solution) is equal to the sum of the partial pressures of each component in the solution:
\[ P_{solution} = X_{water} \cdot P_{0,water} + X_{ethanol} \cdot P_{0,ethanol} \]
Where:
- \(P_{0,water}\) = vapor pressure of pure water = 32.8 mmHg
- \(P_{0,ethanol}\) = vapor pressure of pure ethanol = 40 mmHg

### Step 3: Calculate the partial pressures
Now we can calculate the partial pressures of water and ethanol in the solution:
\[ P_{water} = X_{water} \cdot P_{0,water} = 0.1 \cdot 32.8 = 3.28 \text{ mmHg} \]
\[ P_{ethanol} = X_{ethanol} \cdot P_{0,ethanol} = 0.9 \cdot 40 = 36 \text{ mmHg} \]

### Step 4: Calculate the total vapor pressure of the solution
Now, we can find the total vapor pressure of the solution:
\[ P_{solution} = P_{water} + P_{ethanol} = 3.28 + 36 = 39.28 \text{ mmHg} \]

### Step 5: Round the answer
Rounding to two decimal places, we get:
\[ P_{solution} \approx 39.3 \text{ mmHg} \]

### Final Answer
The vapor pressure of the solution M is approximately **39.3 mmHg**.

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