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)=1.86Kkg mol–1
Freezing point depression constant of ethanol
(Kwaterb)=2.0Kkg mol–1
Boiling point elevation constant of water
(Kwaterb)=0.52Kkg mol–1
Boiling point elevation constant of ethanol
(Kethanolb)=1.2Kkg 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 =18g mol−1
Molecular weight of ethanol =46gmol−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 -
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)=1.86Kkg mol–1
Freezing point depression constant of ethanol
(Kwaterb)=2.0Kkg mol–1
Boiling point elevation constant of water
(Kwaterb)=0.52Kkg mol–1
Boiling point elevation constant of ethanol
(Kethanolb)=1.2Kkg 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 =18g mol−1
Molecular weight of ethanol =46gmol−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 -
The correct option is B 376.2K
Now ethanol is the solute.
Molality of solute = 0.10.9×18×1000=6.17
⇒Δ Tb=6.17×0.52=3.20
⇒Tb=373+3.2=376.2K
376.2K
Now ethanol is the solute.
Molality of solute = 0.10.9×18×1000=6.17
⇒Δ Tb=6.17×0.52=3.20
⇒Tb=373+3.2=376.2K
