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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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer?.

Solutions for 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice 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 Freezing point depression constant of ethanol Boiling point elevation constant of Boiling point elevation constant of ethanol Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol−1Molecular weight of ethanol = 46 g mol−1In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. [JEE 2008]The freezing point of the solution M isa)268.7 Kb)268.5 Kc)234.2 Kd)150.9 KCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.