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MX2 dissociates in M2+ and X- ions in an aqueous solution, with a degree of dissociation (α) of 0.5. The
ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of
freezing point in the absence of ionic dissociation is
    Correct answer is '2'. Can you explain this answer?
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    MX2 dissociates in M2+ and X- ions in an aqueous solution, with a degr...
    Α) of 0.80. This means that 80% of the MX2 molecules dissociate into M2 and X- ions, while the remaining 20% remain as undissociated MX2 molecules.

    The dissociation reaction can be represented as follows:

    MX2 ⇌ M2 + X-

    To calculate the concentration of M2 and X- ions, we need to know the initial concentration of MX2. Let's assume the initial concentration of MX2 is [MX2]0.

    After dissociation, the concentration of undissociated MX2 remaining is 20% of the initial concentration, which can be calculated as follows:

    [MX2] = (20/100) * [MX2]0

    The concentration of M2 and X- ions formed can be calculated using the degree of dissociation (α) and the initial concentration of MX2:

    [M2] = α * [MX2]0

    [X-] = α * [MX2]0

    Substituting the values, we get:

    [M2] = 0.80 * [MX2]0

    [X-] = 0.80 * [MX2]0

    So, 80% of the initial concentration of MX2 dissociates into M2 and X- ions, while 20% remains undissociated.
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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

    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?

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    MX2 dissociates in M2+ and X- ions in an aqueous solution, with a degree of dissociation (α) of 0.5. Theratio of the observed depression of freezing point of the aqueous solution to the value of the depression offreezing point in the absence of ionic dissociation isCorrect answer is '2'. Can you explain this answer?
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    MX2 dissociates in M2+ and X- ions in an aqueous solution, with a degree of dissociation (α) of 0.5. Theratio of the observed depression of freezing point of the aqueous solution to the value of the depression offreezing point in the absence of ionic dissociation 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 MX2 dissociates in M2+ and X- ions in an aqueous solution, with a degree of dissociation (α) of 0.5. Theratio of the observed depression of freezing point of the aqueous solution to the value of the depression offreezing point in the absence of ionic dissociation 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 MX2 dissociates in M2+ and X- ions in an aqueous solution, with a degree of dissociation (α) of 0.5. Theratio of the observed depression of freezing point of the aqueous solution to the value of the depression offreezing point in the absence of ionic dissociation isCorrect answer is '2'. Can you explain this answer?.
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