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25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be
[Kf (H20 ) = 2.0° mol-1 kg, molarity = molality] 
  • a)
    5.00
  • b)
    3.20
  • c)
    1.60
  • d)
    0.80
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1....
(b)
Ag + + Cl- → AgCI
When titration is complete,  gives chocolate coloured precipitate with Ag+.
2Ag+ + → Ag2Cr04
This end point is detected.
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Community Answer
25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1....
(b)
Ag + + Cl- → AgCI
When titration is complete,  gives chocolate coloured precipitate with Ag+.
2Ag+ + → Ag2Cr04
This end point is detected.
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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

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

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

A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vaporizes to form a more disordered gas. When a solulte is present , there is additional contribution to the entropy of the liquid due to increase randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas.Thus , a solute (non volatil e) lowers the vapour pressure of a liquid, and hence a higher booing point of the solution Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent) and the solution . Elevation of B.Pt. Tband depression of F.Pt. Tfof a solution are the colligative properties which depend only on the concentration of particles of the solute, not their identity.For dilute solutions, Tband Tfare proportional to the molality of the solute in the solution.The vaues of Kb and Kf do depend on the properties of the solvent. For liquids, is almost constant . [Troutans Rule , this constant for most of the Unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state) is equal to 90J/mol. ] For solutes undergoing change of molecular state is solution (ionization or association), the observed T values differ from the calculate ones using the above relations. In such situations, the relationships are modified as Where i = Vant Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules.Q.A liquid possessing which of the following characteristics will be most suitable for determining the molecular mass of a compound by cryoscopic measurements?

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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25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be[Kf(H20 ) = 2.0° mol-1kg, molarity = molality]a)5.00b)3.20c)1.60d)0.80Correct answer is option 'B'. Can you explain this answer?
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25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be[Kf(H20 ) = 2.0° mol-1kg, molarity = molality]a)5.00b)3.20c)1.60d)0.80Correct 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 25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be[Kf(H20 ) = 2.0° mol-1kg, molarity = molality]a)5.00b)3.20c)1.60d)0.80Correct 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 25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be[Kf(H20 ) = 2.0° mol-1kg, molarity = molality]a)5.00b)3.20c)1.60d)0.80Correct answer is option 'B'. Can you explain this answer?.
Solutions for 25 mL of an aqueous solu tion of KCI was fo und to require 20 mL of 1.0 M AgN03 solution when titrated using K2Cr04 as an indicator. Thus, d epression in freezing point of KCI solution with 100% ionisation will be[Kf(H20 ) = 2.0° mol-1kg, molarity = molality]a)5.00b)3.20c)1.60d)0.80Correct answer is option 'B'. 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.
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