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. Application 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. 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 `(e_(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 pressusre 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 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 solution M, which is a mixture of ethanol and water with a mole fraction of ethanol (X_ethanol) equal to 0.9, we will follow these steps:

### Step 1: Determine the mole fraction of water
The mole fraction of water (X_water) can be calculated using the relation:
\[ X_{water} = 1 - X_{ethanol} \]
Substituting the given value:
\[ X_{water} = 1 - 0.9 = 0.1 \]

### Step 2: Identify the solute and solvent
Since the mole fraction of ethanol is greater than that of water, we conclude that ethanol is the solvent and water is the solute.

### Step 3: Calculate the molality (m)
Molality (m) can be calculated using the formula:
\[ m = \frac{X_{solute}}{X_{solvent}} \times \frac{1000}{M_{solvent}} \]
Where:
- \(X_{solute} = X_{water} = 0.1\)
- \(X_{solvent} = X_{ethanol} = 0.9\)
- \(M_{solvent} = M_{ethanol} = 46 \, g/mol\)

Substituting the values:
\[ m = \frac{0.1}{0.9} \times \frac{1000}{46} \]
Calculating this gives:
\[ m = \frac{0.1}{0.9} \times 21.7391 \approx 2.415 \, mol/kg \]

### Step 4: Calculate the freezing point depression (ΔTf)
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \times m \]
Where:
- \(K_f\) for ethanol = 2.0 K kg/mol
- \(m \approx 2.415 \, mol/kg\)

Substituting the values:
\[ \Delta T_f = 2.0 \times 2.415 \approx 4.83 \, K \]

### Step 5: Calculate the freezing point of the solution
The freezing point of the solution can be calculated using the formula:
\[ T_f = T_f^0 - \Delta T_f \]
Where:
- \(T_f^0\) (freezing point of pure ethanol) = 155.7 K

Substituting the values:
\[ T_f = 155.7 - 4.83 \approx 150.87 \, K \]

### Final Answer
The freezing point of solution M is approximately **150.9 K**.

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