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Passage III
A solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.
Q.
van’t Hoff factor (i) is
- a)2.0
- b)1.5
- c)1.85
- d)0.85
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in ...
From ΔT, of cane sugar, K, can be determined. 
For NaCI solution,
Most Upvoted Answer
Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in ...
Explanation:
The van't Hoff factor (i) is a measure of the number of particles that a solute molecule dissociates into when it is dissolved in a solvent. It is used to calculate the freezing point depression or boiling point elevation of a solution.
To calculate the van't Hoff factor (i), we need to compare the experimental freezing point depression of the solution with the expected freezing point depression based on the concentration of the solute.
Calculating the expected freezing point depression:
The expected freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m
Where:
ΔTf = Freezing point depression
Kf = Cryoscopic constant (freezing point depression constant) of the solvent (water in this case)
m = molality of the solute in the solution
The molality (m) can be calculated using the formula:
m = moles of solute / mass of solvent (in kg)
For cane sugar (C12H22O11):
Molar mass of cane sugar (C12H22O11) = 342.3 g/mol
moles of cane sugar (C12H22O11) = mass of cane sugar (C12H22O11) / molar mass of cane sugar (C12H22O11)
For NaCl:
Molar mass of NaCl = 58.44 g/mol
moles of NaCl = mass of NaCl / molar mass of NaCl
Calculating the experimental freezing point depression:
The experimental freezing point depression (ΔTf) is given in the problem:
For cane sugar solution: ΔTf = -0.037°C
For NaCl solution: ΔTf = -0.342°C
Calculating the van't Hoff factor (i):
The van't Hoff factor (i) can be calculated using the formula:
i = ΔTf (experimental) / ΔTf (expected)
For cane sugar solution:
ΔTf (expected) = Kf * m (cane sugar)
i (cane sugar) = ΔTf (experimental) / ΔTf (expected)
For NaCl solution:
ΔTf (expected) = Kf * m (NaCl)
i (NaCl) = ΔTf (experimental) / ΔTf (expected)
Calculating the van't Hoff factor (i) for cane sugar solution:
First, calculate the moles of cane sugar:
mass of cane sugar = 0.684 g
moles of cane sugar = 0.684 g / 342.3 g/mol
Next, calculate the molality of the solution:
mass of water = 100 g
mass of water = 100 g / 1000 g/kg
m (cane sugar) = moles of cane sugar / mass of water
Now, calculate the expected freezing point depression:
ΔTf (expected) = Kf * m (cane sugar)
Finally, calculate the van't Hoff factor:
i (cane sugar) = ΔTf (experimental) / ΔTf (expected)
Calculating the van't Hoff factor (i) for NaCl solution:
First, calculate the moles of NaCl:
mass of NaCl =
The van't Hoff factor (i) is a measure of the number of particles that a solute molecule dissociates into when it is dissolved in a solvent. It is used to calculate the freezing point depression or boiling point elevation of a solution.
To calculate the van't Hoff factor (i), we need to compare the experimental freezing point depression of the solution with the expected freezing point depression based on the concentration of the solute.
Calculating the expected freezing point depression:
The expected freezing point depression (ΔTf) can be calculated using the formula:
ΔTf = Kf * m
Where:
ΔTf = Freezing point depression
Kf = Cryoscopic constant (freezing point depression constant) of the solvent (water in this case)
m = molality of the solute in the solution
The molality (m) can be calculated using the formula:
m = moles of solute / mass of solvent (in kg)
For cane sugar (C12H22O11):
Molar mass of cane sugar (C12H22O11) = 342.3 g/mol
moles of cane sugar (C12H22O11) = mass of cane sugar (C12H22O11) / molar mass of cane sugar (C12H22O11)
For NaCl:
Molar mass of NaCl = 58.44 g/mol
moles of NaCl = mass of NaCl / molar mass of NaCl
Calculating the experimental freezing point depression:
The experimental freezing point depression (ΔTf) is given in the problem:
For cane sugar solution: ΔTf = -0.037°C
For NaCl solution: ΔTf = -0.342°C
Calculating the van't Hoff factor (i):
The van't Hoff factor (i) can be calculated using the formula:
i = ΔTf (experimental) / ΔTf (expected)
For cane sugar solution:
ΔTf (expected) = Kf * m (cane sugar)
i (cane sugar) = ΔTf (experimental) / ΔTf (expected)
For NaCl solution:
ΔTf (expected) = Kf * m (NaCl)
i (NaCl) = ΔTf (experimental) / ΔTf (expected)
Calculating the van't Hoff factor (i) for cane sugar solution:
First, calculate the moles of cane sugar:
mass of cane sugar = 0.684 g
moles of cane sugar = 0.684 g / 342.3 g/mol
Next, calculate the molality of the solution:
mass of water = 100 g
mass of water = 100 g / 1000 g/kg
m (cane sugar) = moles of cane sugar / mass of water
Now, calculate the expected freezing point depression:
ΔTf (expected) = Kf * m (cane sugar)
Finally, calculate the van't Hoff factor:
i (cane sugar) = ΔTf (experimental) / ΔTf (expected)
Calculating the van't Hoff factor (i) for NaCl solution:
First, calculate the moles of NaCl:
mass of NaCl =
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Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer?
Question Description
Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. 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 Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. 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 Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer?.
Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. 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 Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. 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 Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer?.
Solutions for Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in 100 g water freezes at - 0.037° C. A solution containing 0.585 g of NaCI in 100 g water freezes at - 0.342° C.Q.van’t Hoff factor (i) isa)2.0b)1.5c)1.85d)0.85Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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