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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.
Apparent molecular weight of NaCI is
- a)58.5 g mol-1
- b)29.25 g mol-1
- c)31.6 g mol-1
- d)117.0 g mol-1
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,
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Passage IIIA solution containing 0.684 g of cane sugar (C12H22011) in ...
°C. Calculate the molality of the solution and the freezing point depression constant (Kf) for water.
To calculate the molality of the solution, we need to first find the moles of cane sugar in the solution.
The molar mass of cane sugar (C12H22011) can be calculated as follows:
C = 12.01 g/mol * 12 = 144.12 g/mol
H = 1.01 g/mol * 22 = 22.22 g/mol
O = 16.00 g/mol * 11 = 176.00 g/mol
Total molar mass = 144.12 + 22.22 + 176.00 = 342.34 g/mol
To find the moles of cane sugar, we divide the mass of cane sugar by its molar mass:
moles of cane sugar = 0.684 g / 342.34 g/mol = 0.001999 mol
The molality of the solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms:
molality = 0.001999 mol / 0.1 kg = 0.01999 mol/kg
Next, we need to calculate the freezing point depression constant (Kf) for water. This value is a constant that depends on the solvent being used.
For water, the value of Kf is 1.86 °C/m.
Therefore, the molality of the solution is 0.01999 mol/kg and the freezing point depression constant (Kf) for water is 1.86 °C/m.
To calculate the molality of the solution, we need to first find the moles of cane sugar in the solution.
The molar mass of cane sugar (C12H22011) can be calculated as follows:
C = 12.01 g/mol * 12 = 144.12 g/mol
H = 1.01 g/mol * 22 = 22.22 g/mol
O = 16.00 g/mol * 11 = 176.00 g/mol
Total molar mass = 144.12 + 22.22 + 176.00 = 342.34 g/mol
To find the moles of cane sugar, we divide the mass of cane sugar by its molar mass:
moles of cane sugar = 0.684 g / 342.34 g/mol = 0.001999 mol
The molality of the solution is calculated by dividing the moles of solute by the mass of the solvent in kilograms:
molality = 0.001999 mol / 0.1 kg = 0.01999 mol/kg
Next, we need to calculate the freezing point depression constant (Kf) for water. This value is a constant that depends on the solvent being used.
For water, the value of Kf is 1.86 °C/m.
Therefore, the molality of the solution is 0.01999 mol/kg and the freezing point depression constant (Kf) for water is 1.86 °C/m.
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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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct answer is option 'C'. Can you explain this answer?
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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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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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Here you can find the meaning 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 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.Apparent molecular weight of NaCI isa)58.5 g mol-1b)29.25 g mol-1c)31.6 g mol-1d)117.0 g mol-1Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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