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 freezing point 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 the components
Given that the mole fraction of ethanol in the mixture is \( 0.9 \), we can determine the mole fraction of water as follows:
\[ X_{\text{water}} = 1 - X_{\text{ethanol}} = 1 - 0.9 = 0.1 \]

### Step 2: Calculate the number of moles of ethanol and water
Assuming we have 1 mole of the solution:
- Moles of ethanol \( (n_e) = 0.9 \)
- Moles of water \( (n_w) = 0.1 \)

### Step 3: Calculate the mass of water
Using the molecular weight of water \( (M_w = 18 \, \text{g/mol}) \):
\[ \text{Mass of water} = n_w \times M_w = 0.1 \, \text{mol} \times 18 \, \text{g/mol} = 1.8 \, \text{g} \]

### Step 4: Convert mass of water to kilograms
\[ \text{Mass of water in kg} = \frac{1.8 \, \text{g}}{1000} = 0.0018 \, \text{kg} \]

### Step 5: Calculate the molality of the solution
Molality \( (m) \) is defined as the number of moles of solute per kilogram of solvent. Here, water is the solvent:
\[ m = \frac{n_e}{\text{mass of water in kg}} = \frac{0.9}{0.0018} = 500 \, \text{mol/kg} \]

### Step 6: Calculate the freezing point depression
Using the freezing point depression constant of water \( (K_f^{\text{water}} = 1.86 \, \text{K kg/mol}) \):
\[ \Delta T_f = K_f \times m = 1.86 \, \text{K kg/mol} \times 500 \, \text{mol/kg} = 930 \, \text{K} \]

### Step 7: Calculate the new freezing point of the solution
The standard freezing point of water is \( 273 \, \text{K} \). Therefore, the freezing point of the solution \( T_f \) is:
\[ T_f = T_f^{\text{water}} - \Delta T_f = 273 \, \text{K} - 930 \, \text{K} = -657 \, \text{K} \]
(Note: This result indicates a calculation error in the molality step, as the molality is too high for a typical solution. Let's correct this.)

### Corrected Step 5: Calculate the molality of the solution
Revisiting the molality calculation:
\[ m = \frac{n_w}{\text{mass of ethanol in kg}} = \frac{0.1}{0.0414} \approx 2.42 \, \text{mol/kg} \]

### Corrected Step 6: Calculate the freezing point depression
Using the corrected molality:
\[ \Delta T_f = K_f \times m = 1.86 \, \text{K kg/mol} \times 2.42 \, \text{mol/kg} \approx 4.5 \, \text{K} \]

### Corrected Step 7: Calculate the new freezing point of the solution
\[ T_f = 273 \, \text{K} - 4.5 \, \text{K} \approx 268.5 \, \text{K} \]

### Final Result
The freezing point of the solution \( M \) is approximately \( 268.5 \, \text{K} \).

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