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The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m is
    Correct answer is '2'. Can you explain this answer?
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    The freezing point of an aqueous solution of KCN containing 0.2 mole/...
    Depression in freezing point - It is the decrease in the temperature at which substance freezes, when a small amount of non-volatile substance is added into it
    ΔTf = Kf × m
    For KCN solution
    0.80 = Kf × 0.2 × 2 ...(1)
    Final effective molality = 0.2 − 0.1 m + 0.1 + 0.2 = 0.5 − 0.1 m
    Now, 0.60 = Kf × (0.5 − 0.1 m) ...(2)
    From (1) & (2) we get,
    m = 2
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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 freezing point of the solution M is

    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?

    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

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    The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer?
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    The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '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 The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '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 The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer?.
    Solutions for The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '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 The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer?, a detailed solution for The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer? has been provided alongside types of The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The freezing point of an aqueous solution of KCN containing 0.2 mole/kg water was −0.80°C. On adding 0.1 of Hg(CN)2 in the solution containing 1 kg of water, the freezing point of the solution was −0.6°C. Assuming that the complex is formed according to the following equation is the limiting reactant, the value of m isCorrect answer is '2'. Can you explain this answer? tests, examples and also practice JEE tests.
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