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JEE Advance 2008
Paper-1
Question
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-today 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 $\left(\mathrm{K}_{\mathrm{f}}^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(\mathrm{K}_{\mathrm{f}}^{\text {ethanol }}\right)=2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Boiling point elevation constant of water $\left(\mathrm{K}_{\mathrm{b}}^{\text {water }}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Boiling point elevation constant of ethanol $\left(\mathrm{K}_{\mathrm{b}}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=32.8 \mathrm{~mm} \mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm} \mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$

In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and nondissociative.

The vapour pressure of the solution M is
Select the correct option:
A
39.3 mm Hg
B
36.0 mm Hg
C
29.5 mm Hg
D
28.8 mm
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
P = 0.9 × 40 = 36 mm Hg
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