A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Water is added to the solution M such that the fraction of water in the solution becomes 0.9. The boiling point of this solution is :

380.4

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376.2

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375.5

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354.7

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Solution

The correct option is B $376.2$ K
The expression for the elevation in the boiling point of the solution is
$Δ T_{b} = K_{b} \times m$ ......(i)

Here, $Δ T_{b}$ is the elevation in the boiling point, $K_{b}$ is the boiling point elevation constant of water
and m is the molality of the solution.

The mole fraction of water is $0.9$ and the mole fraction of ethanol is $0.1$
Thus water acts as a solvent.

The molality of a solution is the number of moles of ethanol (solute) present in 1 kg of water (solvent).
Molar mass of water = $18$ gram/mole.

Hence, the molality of the solution is
$m = \frac{0.1}{0.9 \times 18} \times 1000$ m

Substitute values in equation (i)
The elevation in boiling point is $Δ T_{b} = 0.52 \times \frac{0.1}{0.9 \times 18} \times 1000 = 3.2 K$

The boiling point of pure water is 373 K.

Boiling point of solution $= 373 + 3.2 = 376.2 K$

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Similar questions

Q. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Then freezing point of the solution M is :

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. 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 ethanol and water. The mole fraction of ethanol in the mixture is 0.9.
Given, freezing point depression constant of water
$(K_{f}^{w a t e r}) = 1.86 K k g m o l^{- 1}$
Freezing point depression constant of ethanol
$(K_{b}^{w a t e r}) = 2.0 K k g m o l^{- 1}$
Boiling point elevation constant of water
$(K_{b}^{w a t e r}) = 0.52 K k g m o l^{- 1}$
Boiling point elevation constant of ethanol
$(K_{b}^{e t h a n o l}) = 1.2 K k g m o l^{- 1}$
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 = 351.5K
Vapour pressure of pure water = 32.8 mm Hg
Vapour pressure of pure ethanol = 40 mm Hg
Molecular weight of water $= 18 g m o l^{- 1}$
Molecular weight of ethanol $= 46 g m o l^{- 1}$
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 point of this solution is -

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Q. The vapour pressure of a solution having 0.9 mole fraction of ethanol in ethanol-water is:

Q. The vapour pressure of a solution having 0.9 mole fraction of water in ethanol water is :

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