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An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb /Kf = 0.3, then AB
- a)is 100% ionised at the freezing point of solution
- b)behaves as non-electrolyte at the freezing point of solution
- c)forms dimer at the freezing point of solution
- d)forms trimer at the freezing point of solution
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An aqueous solution of a solute AB has boiling point of 101.08° C...
Thus, AS behaves as a non-electrolyte at the freezing point of solution (remains non-ionised).
Most Upvoted Answer
An aqueous solution of a solute AB has boiling point of 101.08° C...
Explanation:
Boiling Point Elevation:
- The boiling point elevation is given by the formula: ΔTb = i * Kb * m, where i is the van't Hoff factor, Kb is the ebullioscopic constant, and m is the molality of the solution.
- Since AB is 100% ionized at the boiling point, i = 2 (assuming it forms two ions when dissociated).
- Given that the boiling point elevation is 101.08 C, we can calculate the molality of the solution.
Freezing Point Depression:
- The freezing point depression is given by the formula: ΔTf = i * Kf * m, where i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality of the solution.
- Given that the freezing point depression is -1.80 C, we can calculate the molality of the solution.
Analysis:
- Since AB is 100% ionized at the boiling point, it behaves as an electrolyte at that temperature.
- However, at the freezing point, the solution behaves as a non-electrolyte because it freezes at -1.80 C, indicating that the solute does not dissociate into ions at that temperature.
Therefore, option B is correct: AB behaves as a non-electrolyte at the freezing point of the solution.
Boiling Point Elevation:
- The boiling point elevation is given by the formula: ΔTb = i * Kb * m, where i is the van't Hoff factor, Kb is the ebullioscopic constant, and m is the molality of the solution.
- Since AB is 100% ionized at the boiling point, i = 2 (assuming it forms two ions when dissociated).
- Given that the boiling point elevation is 101.08 C, we can calculate the molality of the solution.
Freezing Point Depression:
- The freezing point depression is given by the formula: ΔTf = i * Kf * m, where i is the van't Hoff factor, Kf is the cryoscopic constant, and m is the molality of the solution.
- Given that the freezing point depression is -1.80 C, we can calculate the molality of the solution.
Analysis:
- Since AB is 100% ionized at the boiling point, it behaves as an electrolyte at that temperature.
- However, at the freezing point, the solution behaves as a non-electrolyte because it freezes at -1.80 C, indicating that the solute does not dissociate into ions at that temperature.
Therefore, option B is correct: AB behaves as a non-electrolyte at the freezing point of the solution.
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An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb/Kf = 0.3, then ABa)is 100% ionised at the freezing point of solutionb)behaves as non-electrolyte at the freezing point of solutionc)forms dimer at the freezing point of solutiond)forms trimer at the freezing point of solutionCorrect answer is option 'B'. Can you explain this answer?
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An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb/Kf = 0.3, then ABa)is 100% ionised at the freezing point of solutionb)behaves as non-electrolyte at the freezing point of solutionc)forms dimer at the freezing point of solutiond)forms trimer at the freezing point of solutionCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb/Kf = 0.3, then ABa)is 100% ionised at the freezing point of solutionb)behaves as non-electrolyte at the freezing point of solutionc)forms dimer at the freezing point of solutiond)forms trimer at the freezing point of solutionCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb/Kf = 0.3, then ABa)is 100% ionised at the freezing point of solutionb)behaves as non-electrolyte at the freezing point of solutionc)forms dimer at the freezing point of solutiond)forms trimer at the freezing point of solutionCorrect answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice An aqueous solution of a solute AB has boiling point of 101.08° C and freezes at -1 .80 °C . AB is found to be 100% ionised at boiling point. If Kb/Kf = 0.3, then ABa)is 100% ionised at the freezing point of solutionb)behaves as non-electrolyte at the freezing point of solutionc)forms dimer at the freezing point of solutiond)forms trimer at the freezing point of solutionCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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