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The molal freezing point depression constant for benzene (C6H6) is 4.90 K kg mol-1. Selenium exists as a polymer of the type Sex. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than that of pure benzene. The molecular formula of selenium is
(Atomic mass of Se = 78.8 g mol-1)
  • a)
    Se8
  • b)
    Se6
  • c)
    Se4
  • d)
    Se2
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The molal freezing point depression constant for benzene (C6H6) is 4.9...
Mass of selenium (WB) = 3.26 g
Mass of benzene (WA) = 226 g = 0.226 kg
Depression in freezing point 
Molar depression constant = 4.9 K kg mol-1
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Community Answer
The molal freezing point depression constant for benzene (C6H6) is 4.9...
To calculate the molality (m) of the solution, we need to find the moles of selenium and the mass of benzene.

First, let's find the moles of selenium:
Molar mass of selenium (Se) = 78.96 g/mol
moles of selenium = mass of selenium / molar mass of selenium
moles of selenium = 3.26 g / 78.96 g/mol
moles of selenium = 0.0413 mol

Next, let's find the mass of benzene:
mass of benzene = 226 g

Now, let's calculate the molality (m):
molality (m) = moles of solute / mass of solvent (in kg)
mass of solvent = mass of benzene / 1000 (to convert from grams to kilograms)
mass of solvent = 226 g / 1000
mass of solvent = 0.226 kg

molality (m) = 0.0413 mol / 0.226 kg
molality (m) ≈ 0.183 K kg mol-1

The observed freezing point depression is 0.112 K.

Now, we can use the formula for freezing point depression to calculate the molal freezing point constant (Kf):
∆Tf = Kf * m
Kf = ∆Tf / m
Kf = 0.112 K / 0.183 K kg mol-1
Kf ≈ 0.612 K kg mol-1

Therefore, the molal freezing point depression constant for selenium in benzene is approximately 0.612 K kg mol-1.
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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

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The molal freezing point depression constant for benzene (C6H6) is 4.90 K kg mol-1. Selenium exists as a polymer of the type Sex. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than that of pure benzene. The molecular formula of selenium is(Atomic mass of Se = 78.8 g mol-1)a)Se8b)Se6c)Se4d)Se2Correct answer is option 'A'. Can you explain this answer?
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The molal freezing point depression constant for benzene (C6H6) is 4.90 K kg mol-1. Selenium exists as a polymer of the type Sex. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than that of pure benzene. The molecular formula of selenium is(Atomic mass of Se = 78.8 g mol-1)a)Se8b)Se6c)Se4d)Se2Correct answer is option 'A'. 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 molal freezing point depression constant for benzene (C6H6) is 4.90 K kg mol-1. Selenium exists as a polymer of the type Sex. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than that of pure benzene. The molecular formula of selenium is(Atomic mass of Se = 78.8 g mol-1)a)Se8b)Se6c)Se4d)Se2Correct answer is option 'A'. 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 molal freezing point depression constant for benzene (C6H6) is 4.90 K kg mol-1. Selenium exists as a polymer of the type Sex. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower than that of pure benzene. The molecular formula of selenium is(Atomic mass of Se = 78.8 g mol-1)a)Se8b)Se6c)Se4d)Se2Correct answer is option 'A'. Can you explain this answer?.
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