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.
Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The boiling of the solution is

Options

A. 380.4 K
B. 376.2 K
C. 375.5 K
D. 354.7 K

Explanation

To find the boiling point of the solution after adding water such that the mole fraction of water becomes 0.9, we will follow these steps:

### Step 1: Determine the Mole Fraction of the Solute
Given that the mole fraction of water (solvent) is 0.9, we can find the mole fraction of ethanol (solute) using the relationship:

\[ X_{solute} = 1 - X_{solvent} \]

Substituting the value:

\[ X_{solute} = 1 - 0.9 = 0.1 \]

### Step 2: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. To find the molality, we need to use the mole fraction and the molecular weight of the solvent (water).

1. Assume we have 1 kg of water. The number of moles of water can be calculated as:

\[ \text{Moles of water} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol} \]

2. The number of moles of ethanol can be calculated from the mole fraction:

\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \implies 0.1 = \frac{n_{solute}}{n_{solute} + 55.56} \]

Let \( n_{solute} \) be the number of moles of ethanol. Rearranging gives:

\[ 0.1(n_{solute} + 55.56) = n_{solute} \]

This leads to:

\[ 0.1n_{solute} + 5.556 = n_{solute} \implies 0.9n_{solute} = 5.556 \implies n_{solute} = \frac{5.556}{0.9} \approx 6.17 \text{ mol} \]

3. Now, we can calculate the molality:

\[ m = \frac{n_{solute}}{\text{mass of solvent in kg}} = \frac{6.17 \text{ mol}}{1 \text{ kg}} = 6.17 \text{ mol/kg} \]

### Step 3: Calculate the Boiling Point Elevation
Using the formula for boiling point elevation:

\[ \Delta T_b = K_b \cdot m \]

Where:
- \( K_b \) for water = 0.52 K kg/mol
- \( m = 6.17 \text{ mol/kg} \)

Substituting the values:

\[ \Delta T_b = 0.52 \cdot 6.17 \approx 3.20 \text{ K} \]

### Step 4: Calculate the New Boiling Point
The boiling point of pure water is 373 K. Therefore, the boiling point of the solution is:

\[ T_b = T_{b0} + \Delta T_b = 373 \text{ K} + 3.20 \text{ K} = 376.2 \text{ K} \]

### Final Answer
The boiling point of the solution is **376.2 K**.

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