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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-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 (Kwaterf)=1.86K kg mol−1 Freezing point depression constant of ethanol (Kwaterb)=2.0K kg mol−1 Boiling point elevation constant of water (Kwaterb)=0.52K kg mol−1 Boiling point elevation constant of ethanol (Kethanolb)=1.2K 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 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 mol−1 Molecular 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. The vapour pressure of the solution M is

A

39.3 mm Hg

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B

36.0 mm Hg

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C

29.5 mm Hg

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D

28.8 mm Hg

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Solution

## The correct option is A 39.3 mm Hg Vapour pressure = p(H2O)+p(ethanol) Vapour pressure=p(H2O)+p(ethanol) =32.8×0.1+40×0.9 =3.28+36 =39.28 mm

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