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)^(water)=1.86 K kg mol^(-1))`
Freezing point depression constant to ethanol
`(K_(f)^(ethanol))=2.0 K kg mol^(-1))`
Boiling point elevation constant of water
`(K_(b)^(water))=0.52 K kg mol^(-1))`
Boiling point elevation constant of ethanol
`(K_(b)^(ethanol))=1.2 K 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 = `351.5 K `
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 anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
The freezing point of the solution `M` is
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)^(water)=1.86 K kg mol^(-1))`
Freezing point depression constant to ethanol
`(K_(f)^(ethanol))=2.0 K kg mol^(-1))`
Boiling point elevation constant of water
`(K_(b)^(water))=0.52 K kg mol^(-1))`
Boiling point elevation constant of ethanol
`(K_(b)^(ethanol))=1.2 K 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 = `351.5 K `
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 anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
The freezing point of the solution `M` is
Given: Freezing point depression constant of water
`(K_(f)^(water)=1.86 K kg mol^(-1))`
Freezing point depression constant to ethanol
`(K_(f)^(ethanol))=2.0 K kg mol^(-1))`
Boiling point elevation constant of water
`(K_(b)^(water))=0.52 K kg mol^(-1))`
Boiling point elevation constant of ethanol
`(K_(b)^(ethanol))=1.2 K 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 = `351.5 K `
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 anwering the following questions consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.
The freezing point of the solution `M` is
A
`268.7 K`
B
`268.5 K`
C
`234.2 K`
D
`150.9 K`
Text Solution
Verified by Experts
The correct Answer is:
D
Solution M is mixture of ethanol and water, Mole fraction of ethanol is 0.9 (Solvent is `C_(2)H_(5)OH`) Mole fraction water is 0.1 (`H_(2)O` is solute)
Molality of `H_(2)O=(n_(2)xx1000)/(n_(1)M_(1))`
`=(0.1xx1000)/(0.9xx46)=2.415` m
`DeltaT_(f)=K_(f)m=2xx2.2415=4.83`
Freezing point of solution
`= 155.7 - 4.83 = 150.87 K`
Total vapour pressure, `p=p_(A)^(@)xx A`.
Molality of `H_(2)O=(n_(2)xx1000)/(n_(1)M_(1))`
`=(0.1xx1000)/(0.9xx46)=2.415` m
`DeltaT_(f)=K_(f)m=2xx2.2415=4.83`
Freezing point of solution
`= 155.7 - 4.83 = 150.87 K`
Total vapour pressure, `p=p_(A)^(@)xx A`.
Topper's Solved these Questions
SOLUTIONS
DINESH PUBLICATION|Exercise Paragraph 2|1 VideosSOLUTIONS
DINESH PUBLICATION|Exercise Paragraph|6 VideosSOLUTIONS
DINESH PUBLICATION|Exercise OBJECTIVE TYPE MCQs|47 VideosSOLID STATE
DINESH PUBLICATION|Exercise Brain storming|10 VideosSOME BASIC CONCEPTS OF CHEMISTRY
DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos
Similar Questions
Explore conceptually related problems
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. Thus moel fraction of ethanol in the mixture is 0.9 . Given: Freezing point depression constant of water (K_(f)^("water")) = 1.86 K kg mol^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol")) = 2.0 K kg mol^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg mol^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecualr weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In asweering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution M is :
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)^(water)=1.86 K kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=2.0 K kg mol^(-1)) Boiling point elevation constant of water (K_(b)^(water))=0.52 K kg mol^(-1)) Boiling point elevation constant of ethanol (K_(b)^(ethanol))=1.2 K 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 = 351.5 K 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 anwering 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. Thus moel fraction of ethanol in the mixture is 0.9 . Given: Freezing point depression constant of water (K_(f)^("water")) = 1.86 K kg mol^(-1) Freezing point depression constant of ethanol (K_(f)^("ethanol")) = 2.0 K kg mol^(-1) Boiling point elevation constant of water (K_(b)^("water")) = 0.52 K kg mol^(-1) Boiling point elevation constant of ethanol (K_(b)^("ethanol")) = 1.2 K kg mol^(-1) Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure ethanol = 40 mm Hg Molecualr weight of water = 18 g mol^(-1) Molecular weight of ethanol = 46 g mol^(-1) In asweering 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 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)^(water)=1.86 K kg mol^(-1)) Freezing point depression constant to ethanol (K_(f)^(ethanol))=2.0 K kg mol^(-1)) Boiling point elevation constant of water (K_(b)^(water))=0.52 K kg mol^(-1)) Boiling point elevation constant of ethanol (K_(b)^(ethanol))=1.2 K 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 = 351.5 K 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 anwering 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