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What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.?
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Freezing Point Depression Concept
The freezing point depression of a solution can be calculated using the formula:
ΔTf = Kf * m * i
Where:
- ΔTf = change in freezing point
- Kf = freezing point depression constant of the solvent
- m = molality of the solution
- i = van't Hoff factor (degree of dissociation)
Given Data
- Freezing point of benzene (Tf) = 278.4 K
- Molar heat of fusion of benzene = 10.042 U/mol
- Molality of acetic acid solution = 0.4 molal
- Degree of dimerization = 85%
Calculation of Van't Hoff Factor (i)
Since acetic acid (CH3COOH) dimerizes to form (CH3COOH)2, the extent of dimerization affects the van’t Hoff factor:
- Initial moles of acetic acid = 0.4
- Dimerization = 85%
- Moles after dimerization = 0.4 * (1 - 0.85) + 0.4 * 0.85 / 2 = 0.4 * 0.15 + 0.4 * 0.425 = 0.06 + 0.17 = 0.23
Thus, the van’t Hoff factor (i) ≈ 1.73 (because 0.4 moles produce 0.23 effective moles).
Freezing Point Depression Calculation
- Kf for benzene = 5.12 K kg/mol (standard value).
- ΔTf = Kf * m * i = 5.12 * 0.4 * 1.73 ≈ 3.55 K.
Final Freezing Point
- Final freezing point = Initial freezing point - ΔTf
= 278.4 K - 3.55 K ≈ 274.85 K.
Conclusion
The freezing point of the 0.4 molal acetic acid solution in benzene, accounting for 85% dimerization, is approximately 274.85 K.
The freezing point depression of a solution can be calculated using the formula:
ΔTf = Kf * m * i
Where:
- ΔTf = change in freezing point
- Kf = freezing point depression constant of the solvent
- m = molality of the solution
- i = van't Hoff factor (degree of dissociation)
Given Data
- Freezing point of benzene (Tf) = 278.4 K
- Molar heat of fusion of benzene = 10.042 U/mol
- Molality of acetic acid solution = 0.4 molal
- Degree of dimerization = 85%
Calculation of Van't Hoff Factor (i)
Since acetic acid (CH3COOH) dimerizes to form (CH3COOH)2, the extent of dimerization affects the van’t Hoff factor:
- Initial moles of acetic acid = 0.4
- Dimerization = 85%
- Moles after dimerization = 0.4 * (1 - 0.85) + 0.4 * 0.85 / 2 = 0.4 * 0.15 + 0.4 * 0.425 = 0.06 + 0.17 = 0.23
Thus, the van’t Hoff factor (i) ≈ 1.73 (because 0.4 moles produce 0.23 effective moles).
Freezing Point Depression Calculation
- Kf for benzene = 5.12 K kg/mol (standard value).
- ΔTf = Kf * m * i = 5.12 * 0.4 * 1.73 ≈ 3.55 K.
Final Freezing Point
- Final freezing point = Initial freezing point - ΔTf
= 278.4 K - 3.55 K ≈ 274.85 K.
Conclusion
The freezing point of the 0.4 molal acetic acid solution in benzene, accounting for 85% dimerization, is approximately 274.85 K.
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What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.? 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 What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.?.
What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.? 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 What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the freezing point of 0.4molal solution of acetic acid in benzene in which it dimerises to the extent of 85%.freezing point of benzene is 278.4k and its molar heat of fusion is 10.042U/mol.?.
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