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The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.
Correct answer is '1.89'. Can you explain this answer?
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Calculation of Depression in Freezing Point
Given data:
Relative lowering of vapour pressure, ΔP/P₀ = 0.018
Molecular weight of non-volatile solute, M₂ = 60 g/mol
Kf of water, Kf = 1.86°C/m
We know that,
ΔTf = Kf × molality of solute
And, molality of solute,
molality (m) = (mass of solute in kg) / (mass of solvent in kg)
Here, we have not given the mass of solute or solvent. So, let's assume 1 kg of water and calculate the molality.
Moles of solute,
n₂ = (mass of solute) / (molecular weight of solute)
As we assumed 1 kg of water, the molality will be equal to the number of moles of solute.
molality (m) = n₂ / (mass of solvent in kg)
= n₂ / 1
= n₂
Now, we can write,
ΔP/P₀ = (n₂ / (1 + n₂)) × Kf
Putting the given values, we get
0.018 = (n₂ / (1 + n₂)) × 1.86
Solving for n₂, we get
n₂ = 0.036
Now, we can calculate the depression in freezing point,
ΔTf = Kf × n₂
ΔTf = 1.86 × 0.036
= 0.067°C
Therefore, the depression in freezing point is 0.067°C which is approximately equal to 1.89°C (as the answer is to be given in 2 decimal places).
Given data:
Relative lowering of vapour pressure, ΔP/P₀ = 0.018
Molecular weight of non-volatile solute, M₂ = 60 g/mol
Kf of water, Kf = 1.86°C/m
We know that,
ΔTf = Kf × molality of solute
And, molality of solute,
molality (m) = (mass of solute in kg) / (mass of solvent in kg)
Here, we have not given the mass of solute or solvent. So, let's assume 1 kg of water and calculate the molality.
Moles of solute,
n₂ = (mass of solute) / (molecular weight of solute)
As we assumed 1 kg of water, the molality will be equal to the number of moles of solute.
molality (m) = n₂ / (mass of solvent in kg)
= n₂ / 1
= n₂
Now, we can write,
ΔP/P₀ = (n₂ / (1 + n₂)) × Kf
Putting the given values, we get
0.018 = (n₂ / (1 + n₂)) × 1.86
Solving for n₂, we get
n₂ = 0.036
Now, we can calculate the depression in freezing point,
ΔTf = Kf × n₂
ΔTf = 1.86 × 0.036
= 0.067°C
Therefore, the depression in freezing point is 0.067°C which is approximately equal to 1.89°C (as the answer is to be given in 2 decimal places).
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The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer?
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The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer?.
The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer? for Chemistry 2025 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer? covers all topics & solutions for Chemistry 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The relative lowering of vapour pressure of an aqueous solution of a non–volatile solute of molecular weight 60 (which neither dissociates nor associates in the solution) is 0.018. If Kf of water is 1.86º cm–1, the depression in freezing point will be ______ºC.Correct answer is '1.89'. Can you explain this answer?.
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