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
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
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
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
