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 :

268.7

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268.5

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234.2

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150.9

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Solution

The correct option is D $150.9$ K
The expression for the depression in the freezing point of the solution is
$Δ T_{f} = K_{f} \times m$ ......(i)

Here, $Δ T_{f}$ is the depression in the freezing point, $K_{f}$ is the freezing point depression constant of ethanol
and m is the molality of solution.

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

The molality of solution is the number of moles of water (solute) present in 1 kg of ethanol (solvent).
The molar mass of ethanol is $46 \times 10^{- 3} k g / m o l$

Hence, the molality of the solution is
$m = \frac{0.1}{0.9 \times 46} \times 1000$ m

Substitute values in equation (i)
The depression in the freezing point is $Δ T_{f} = 2 \times \frac{0.1}{0.9 \times 46} \times 1000 = 4.83 K$

The freezing point of pure ethanol is 155.7 K.

Freezing point of solution $M = 155.7 - 4.83 = 150.87 K \approx 150.9 K$

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Q. 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 :

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.