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 adn water mixtures 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 `(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 `= 155.7 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 pressure 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 freezing point of the solution M is

Options

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

Explanation

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

### Step 1: Identify the mole fractions
Given that the mole fraction of ethanol (solvent) is \(X_A = 0.9\), we can find the mole fraction of water (solute):
\[ X_B = 1 - X_A = 1 - 0.9 = 0.1 \]

### Step 2: Set up the relationship between moles
Using the mole fractions, we can express the number of moles of solvent (ethanol) and solute (water):
\[ \frac{X_A}{X_B} = \frac{N_A}{N_B} \quad \text{where } N_A \text{ is moles of ethanol and } N_B \text{ is moles of water.} \]
Substituting the values:
\[ \frac{0.9}{0.1} = \frac{N_A}{N_B} \implies 9 = \frac{N_A}{N_B} \implies N_B = \frac{N_A}{9} \]

### Step 3: Express molality
Molality (m) is defined as:
\[ m = \frac{N_B}{\text{mass of solvent (kg)}} \]
We can express \(N_B\) in terms of \(N_A\):
\[ N_B = \frac{N_A}{9} \implies m = \frac{N_A/9}{\text{mass of solvent (kg)}} \]

### Step 4: Calculate mass of solvent
Assuming 1 mole of ethanol (molar mass = 46 g/mol), the mass of ethanol is:
\[ \text{mass of ethanol} = 46 \text{ g} = 0.046 \text{ kg} \]

### Step 5: Calculate molality
Substituting into the molality equation:
\[ m = \frac{(N_A/9)}{0.046} \quad \text{and since } N_A = 1 \text{ mole,} \]
\[ m = \frac{1/9}{0.046} \approx \frac{0.1111}{0.046} \approx 2.41 \text{ mol/kg} \]

### Step 6: Calculate freezing point depression
Using the formula for freezing point depression:
\[ \Delta T_F = K_F \cdot m \]
Where \(K_F\) for ethanol is given as \(2.0 \text{ K kg mol}^{-1}\):
\[ \Delta T_F = 2.0 \cdot 2.41 \approx 4.82 \text{ K} \]

### Step 7: Calculate the new freezing point
The standard freezing point of ethanol is \(155.7 \text{ K}\):
\[ T_F' = T_F - \Delta T_F = 155.7 - 4.82 \approx 150.88 \text{ K} \]

### Final Answer
Thus, the freezing point of solution M is approximately:
\[ \text{Freezing point of solution M} \approx 150.9 \text{ K} \]