Properties such as boiling point, freezing point and vapour, pressure of a pure solvent change Propeties 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 athanol 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 kg "mol"^(-1)`
Boiling point elevation constant of ethanol `(K_(b)^("ethanol"))=1.2 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
tandard boiling point of ethanol =351.5 K
Vapour pressure of pure water =32.8 mmHg
Vapour presure of pure ethanol =40g Hg
Molecular weight of water `=18 g"mol"^(-1)`
Molecules 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.
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 athanol and water. The mole fraction of ethanol in the mixture is 0.9 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: Identify the Mole Fractions
Given that the mole fraction of ethanol (solvent) is 0.9, we can calculate the mole fraction of water (solute):
\[ \text{Mole fraction of water} = 1 - \text{Mole fraction of ethanol} = 1 - 0.9 = 0.1 \]

### Step 2: Determine the Molecular Weights
From the data provided:
- Molecular weight of water = 18 g/mol
- Molecular weight of ethanol = 46 g/mol

### Step 3: Calculate Molality
Molality (m) is defined as the number of moles of solute per kilogram of solvent. We can use the mole fraction to find the molality:
\[ \text{Molality} = \frac{\text{Mole fraction of solute} \times 1000}{\text{Mole fraction of solvent} \times \text{Molecular weight of solvent}} \]
Substituting the values:
\[ \text{Molality} = \frac{0.1 \times 1000}{0.9 \times 46} = \frac{100}{41.4} \approx 2.415 \, \text{mol/kg} \]

### Step 4: Calculate Freezing Point Depression
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \times m \]
Where:
- \(K_f\) for ethanol = 2.0 K kg/mol
- \(m\) = 2.415 mol/kg (calculated in the previous step)

Substituting the values:
\[ \Delta T_f = 2.0 \, \text{K kg/mol} \times 2.415 \, \text{mol/kg} \approx 4.83 \, \text{K} \]

### Step 5: Calculate the New Freezing Point
The standard freezing point of ethanol is 155.7 K. The new freezing point of the solution can be calculated as:
\[ \text{New Freezing Point} = \text{Initial Freezing Point} - \Delta T_f \]
Substituting the values:
\[ \text{New Freezing Point} = 155.7 \, \text{K} - 4.83 \, \text{K} \approx 150.87 \, \text{K} \]

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

---