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Directions: Read the paragraph and answer the following question.
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1
Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1
Boiling point elevation constant of water = 0.52 K kg mol-1
Boiling point elevation constant of 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
Molecular weight of water = 18 g mol-1
Molecular weight of ethanol = 46 g mol-1
In answering the following question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.
    Correct answer is '376.2'. Can you explain this answer?
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    Directions: Read the paragraph and answer the following question.Prop...
    ΔTb = Kb.molality
    = 0.52 x 6.1728
    (T0 - TB0) = 3.20
    Tb = 373 + 3.20
    = 376.2 K
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    PassageProperties 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() = l . 8 6 K kg mol-1Freezing point depression constant of ethanol () = 2.0 K kg mol-1Boiling point elevation constant of water {) = 0.52 K kg mol-1Boiling point elevation constant of ethanol ()=1.2 K kg mol-1Standard 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-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.Q.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

    PassageProperties 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() = l . 8 6 K kg mol-1Freezing point depression constant of ethanol () = 2.0 K kg mol-1Boiling point elevation constant of water {) = 0.52 K kg mol-1Boiling point elevation constant of ethanol ()=1.2 K kg mol-1Standard 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-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.Q.The vapour pressure of the solution M is

    PassageProperties 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() = l . 8 6 K kg mol-1Freezing point depression constant of ethanol () = 2.0 K kg mol-1Boiling point elevation constant of water {) = 0.52 K kg mol-1Boiling point elevation constant of ethanol ()=1.2 K kg mol-1Standard 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-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.Q.The freezing point 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-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 is

    Properties, whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionisation or association hence these properties are also affected. Number of moles of the product is related to degree of ionisation or association by vant Hoff factor ‘i’ given by i = [ 1 + (n – 1) α] for dissociationwhere n is the number of products (ions or molecules) obtained per mole of the reactant & i = for association.Where n is number of reactant particles associated to give 1 mole product.A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb. The solute dimerises in the solution according to the following equation. The degree of association is α.Q.The colligative properties observed will be

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    Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer?
    Question Description
    Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. 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 Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer?.
    Solutions for Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. 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 Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer?, a detailed solution for Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer? has been provided alongside types of Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Directions: Read the paragraph and answer the following question.Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogenous 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 (Kwater) = 1.86 K kg mol-1Freezing point depression constant on ethanol (Kethanol) = 2.0 K kg mol-1Boiling point elevation constant of water = 0.52 K kg mol-1Boiling point elevation constant of ethanol = 1.2 K kg mol-1Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7KStandard 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 question, consider the situations to be ideal dilute solutions and solutes to be non-volatile and non-dissociate.Correct answer is '376.2'. Can you explain this answer? tests, examples and also practice JEE tests.
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