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 f.p. of the solution M is

Options

A. 268.7 K
B. 268.5 K
C. 234.2 K
D. 150.9 K

Explanation

To find the freezing point of the solution M prepared by mixing ethanol and water, we will follow these steps:

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

### Step 2: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. In this case, ethanol is the solute and water is the solvent.

To find the molality, we need to calculate the moles of ethanol and the mass of water in kilograms. We can assume we have 1 mole of the solution for simplicity.

- Moles of ethanol = 0.9 moles (since the mole fraction of ethanol is 0.9)
- Moles of water = 0.1 moles (since the mole fraction of water is 0.1)

Now, we need to convert the moles of water to mass:
\[ \text{Mass of water} = \text{moles} \times \text{molar mass} = 0.1 \, \text{moles} \times 18 \, \text{g/mol} = 1.8 \, \text{g} = 0.0018 \, \text{kg} \]

Now we can calculate the molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.9 \, \text{moles}}{0.0018 \, \text{kg}} = 500 \, \text{mol/kg} \]

### Step 3: Calculate the depression in freezing point (ΔTf)
The depression in freezing point is given by the formula:
\[ \Delta T_f = K_f \times m \]
Where:
- \( K_f \) for water = 1.86 K kg/mol
- \( m = 500 \, \text{mol/kg} \)

Substituting the values:
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 500 \, \text{mol/kg} = 930 \, \text{K} \]

### Step 4: Calculate the freezing point of the solution
The freezing point of pure water is 273 K. Therefore, the freezing point of the solution (Tf_solution) can be calculated as:
\[ T_f^{solution} = T_f^{pure} - \Delta T_f = 273 \, \text{K} - 930 \, \text{K} = -657 \, \text{K} \]

### Conclusion
The freezing point of the solution M is -657 K.