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Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, then how many grams of ethylene glycol (C2H602) must you add to get the freezing point of the solution lowered to -2.8oc ?
(AIEEE 2012)
  • a)
    72 g
  • b)
    93 g
  • c)
    39 g
  • d)
    27 g
Correct answer is option 'B'. Can you explain this answer?
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Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0...
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Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0...
Given data:
Kf for water (Kf) = 1.86 K kg mol⁻¹
Mass of water (m1) = 1.0 kg
Freezing point depression (ΔTf) = -2.8 °C
Molar mass of ethylene glycol (M2) = 62 g mol⁻¹

To find: Mass of ethylene glycol (m2) required to lower the freezing point of water to -2.8 °C

Formula used:
ΔTf = Kf x molality
Molality (m) = moles of solute (n2) / mass of solvent (m1)
Moles of solute (n2) = m2 / M2

Calculation:
From the formula, ΔTf = Kf x molality
molality (m) = n2 / m1

Substitute the given values, we get
-2.8 = 1.86 x (n2 / 1.0)
n2 = (2.8 / 1.86) = 1.50 mol

Molar mass of ethylene glycol (M2) = 62 g mol⁻¹
Mass of ethylene glycol (m2) = n2 x M2
= 1.50 x 62
= 93 g

Therefore, 93 g of ethylene glycol must be added to 1.0 kg of water to lower the freezing point to -2.8 °C. Hence, the correct answer is option (b).
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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?

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.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

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Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, then how many grams of ethylene glycol (C2H602) must you add to get the freezing point of the solution lowered to -2.8oc ?(AIEEE 2012)a)72 gb)93 gc)39 gd)27 gCorrect answer is option 'B'. Can you explain this answer?
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Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, then how many grams of ethylene glycol (C2H602) must you add to get the freezing point of the solution lowered to -2.8oc ?(AIEEE 2012)a)72 gb)93 gc)39 gd)27 gCorrect answer is option 'B'. 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 Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, then how many grams of ethylene glycol (C2H602) must you add to get the freezing point of the solution lowered to -2.8oc ?(AIEEE 2012)a)72 gb)93 gc)39 gd)27 gCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, then how many grams of ethylene glycol (C2H602) must you add to get the freezing point of the solution lowered to -2.8oc ?(AIEEE 2012)a)72 gb)93 gc)39 gd)27 gCorrect answer is option 'B'. Can you explain this answer?.
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