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
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.
Water is added to the solution M such that the fraction of water in the solution becomes `0.9`. The boiling point of this solutions is

Options

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

Explanation

To solve the problem, we need to find the new 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 ethanol
Given that the mole fraction of water is 0.9, we can find the mole fraction of ethanol (the solute) using the relation:
\[ \text{Mole fraction of ethanol} = 1 - \text{Mole fraction of water} = 1 - 0.9 = 0.1 \]

### Step 2: Calculate the molality of the solution
Molality (m) can be calculated using the following formula:
\[ m = \frac{\text{Mole fraction of solute} \times 1000}{\text{Mole fraction of solvent} \times \text{Molecular weight of solvent}} \]
Here, the mole fraction of the solute (ethanol) is 0.1, the mole fraction of the solvent (water) is 0.9, and the molecular weight of water is 18 g/mol. Plugging in these values:
\[ m = \frac{0.1 \times 1000}{0.9 \times 18} = \frac{100}{16.2} \approx 6.17 \text{ mol/kg} \]

### Step 3: Calculate the boiling point elevation
The boiling point elevation (ΔT_b) can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Where \(K_b\) for water is given as 0.52 kg/mol. Therefore:
\[ \Delta T_b = 0.52 \times 6.17 \approx 3.20 \text{ K} \]

### Step 4: Determine the new boiling point
The standard boiling point of water is 373 K. To find the new boiling point of the solution, we add the boiling point elevation to the standard boiling point:
\[ T_b(\text{new}) = T_b(\text{initial}) + \Delta T_b = 373 + 3.20 = 376.20 \text{ K} \]

### Final Answer
The new boiling point of the solution is **376.20 K**.

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