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AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a 10.0 molal aqueous solution of AB2 is ________°C.
(Round off to the nearest integer)
[Given: Molal elevation constant of water, Kb = 0.5 K kg mol-1; boiling point of pure water = 100°C]
    Correct answer is '106'. Can you explain this answer?
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    AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a ...
    The boiling point of a solution depends on the concentration of solute particles in the solution. In this case, AB2 dissociates into A2 and B-, so there are three solute particles (A2, B-, and AB2) in the solution.

    A 10.0 molal solution means that there are 10 moles of solute (AB2) per kilogram of solvent (water).

    Since AB2 is 10% dissociated, only 10% of the 10 moles of AB2 will dissociate, giving us 1 mole of A2 and 1 mole of B- in the solution.

    Therefore, the total number of solute particles in the solution is 1 (A2) + 1 (B-) + 1 (AB2) = 3.

    The boiling point elevation of a solution can be calculated using the equation:

    ΔTb = Kb * m * i,

    where ΔTb is the boiling point elevation, Kb is the molal boiling point elevation constant for the solvent (water), m is the molality of the solution, and i is the van't Hoff factor (the number of particles into which the solute dissociates).

    We need to find the boiling point elevation, so we can rearrange the equation to solve for ΔTb:

    ΔTb = Kb * m * i.

    Assuming we have the value for Kb for water, we can plug in the values:

    ΔTb = Kb * 10.0 molal * 3.

    The boiling point elevation will depend on the specific value of Kb for water, which can be found in a reference table.

    Therefore, without the specific value of Kb for water, we cannot determine the boiling point of the 10.0 molal aqueous solution of AB2.
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    Community Answer
    AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a ...
    AB2  ⇌ A+ + 2B-
    ∴ For AB2, n = 3
    i = 1 + (n - 1)α
    = 1 + (3 - 1) x 0.1
    = 1.12
    Now, ΔTb = Kb(im)
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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

    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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    AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a 10.0 molal aqueous solution of AB2 is ________°C.(Round off to the nearest integer)[Given: Molal elevation constant of water, Kb = 0.5 K kg mol-1; boiling point of pure water = 100°C]Correct answer is '106'. Can you explain this answer?
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    AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a 10.0 molal aqueous solution of AB2 is ________°C.(Round off to the nearest integer)[Given: Molal elevation constant of water, Kb = 0.5 K kg mol-1; boiling point of pure water = 100°C]Correct answer is '106'. 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 AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a 10.0 molal aqueous solution of AB2 is ________°C.(Round off to the nearest integer)[Given: Molal elevation constant of water, Kb = 0.5 K kg mol-1; boiling point of pure water = 100°C]Correct answer is '106'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for AB2 is 10% dissociated in water to A2+ and B-. The boiling point of a 10.0 molal aqueous solution of AB2 is ________°C.(Round off to the nearest integer)[Given: Molal elevation constant of water, Kb = 0.5 K kg mol-1; boiling point of pure water = 100°C]Correct answer is '106'. Can you explain this answer?.
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