Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example 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 "mol"^(-1)`
Freezing point depression constant of ethanol `(K_(f)^("ethonal")) = 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)^("ethonal")) = 1.2 K kg mol^(-1)`
Standard freezing point of water `= 273 K`
Standard freezing point of ethonal `= 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 ethonal `= 40 mm Hg`
Molecular weight of water `= 18 g "mol"^(-1)`
Molecular weight of ethonal `= 45 g"mol"^(-1)`
In answering the following questions, consider the solution to be ideal ideal solutions and solutes to be non-volatile and non-dissociative.
Water is added to the solution M such lthat the molecules fraction of water in t he solution becomes `0.9`. The boiling point of this solution is :

Options

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

Explanation

To solve the problem step-by-step, we need to determine the boiling point of the solution after adding water such that the mole fraction of water becomes 0.9.

### Step 1: Determine the mole fraction of the solute (ethanol)
Given that the mole fraction of ethanol in the mixture is 0.9, we can find the mole fraction of the solute (ethanol) as follows:

\[ \text{Mole fraction of solute (ethanol)} = 1 - \text{Mole fraction of solvent (water)} = 1 - 0.9 = 0.1 \]
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