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The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will be
- a)–7.44ºC
- b)–5.58ºC
- c)–3.72ºC
- d)–2.79ºC
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The freezing point of 0.1 M solution of glucose is -1.860C. If an equa...
-1.240C
To solve this problem, we can use the formula:
ΔTf = Kf × m
where ΔTf is the change in freezing point, Kf is the freezing point depression constant (for water, Kf = 1.86 C/m), and m is the molality of the solution (moles of solute per kg of solvent).
For the first solution, we have:
ΔTf1 = Kf × m1
where m1 = 0.1 mol/kg.
Substituting values, we get:
-1.86 = 1.86 × 0.1 × i1
where i1 is the van't Hoff factor, which is 1 for glucose.
Solving for i1, we get:
i1 = -1.86/1.86/0.1 = -10
This is obviously incorrect, since the van't Hoff factor cannot be negative. The error probably comes from a mistake in the units (e.g. using grams instead of kilograms).
Assuming that the molality of the first solution is actually 0.1 mol/kg, we can proceed as follows:
ΔTf1 = Kf × m1 = 1.86 × 0.1 = 0.186 C
Now we add an equal volume of 0.3 M glucose solution, which has a molality of 0.3 mol/kg. The total molality of the mixture is:
m2 = (m1V1 + m2V2)/(V1 + V2)
where V1 and V2 are the volumes of the two solutions, which we assume to be equal. Substituting values, we get:
m2 = (0.1 × 1 + 0.3 × 1)/(1 + 1) = 0.2 mol/kg
The change in freezing point for the mixture is:
ΔTf2 = Kf × m2 = 1.86 × 0.2 = 0.372 C
The freezing point of the mixture is:
Tf2 = Tf1 - ΔTf2 = -1.86 - 0.372 = -1.488 C
Therefore, the freezing point of the mixture is -1.488C.
To solve this problem, we can use the formula:
ΔTf = Kf × m
where ΔTf is the change in freezing point, Kf is the freezing point depression constant (for water, Kf = 1.86 C/m), and m is the molality of the solution (moles of solute per kg of solvent).
For the first solution, we have:
ΔTf1 = Kf × m1
where m1 = 0.1 mol/kg.
Substituting values, we get:
-1.86 = 1.86 × 0.1 × i1
where i1 is the van't Hoff factor, which is 1 for glucose.
Solving for i1, we get:
i1 = -1.86/1.86/0.1 = -10
This is obviously incorrect, since the van't Hoff factor cannot be negative. The error probably comes from a mistake in the units (e.g. using grams instead of kilograms).
Assuming that the molality of the first solution is actually 0.1 mol/kg, we can proceed as follows:
ΔTf1 = Kf × m1 = 1.86 × 0.1 = 0.186 C
Now we add an equal volume of 0.3 M glucose solution, which has a molality of 0.3 mol/kg. The total molality of the mixture is:
m2 = (m1V1 + m2V2)/(V1 + V2)
where V1 and V2 are the volumes of the two solutions, which we assume to be equal. Substituting values, we get:
m2 = (0.1 × 1 + 0.3 × 1)/(1 + 1) = 0.2 mol/kg
The change in freezing point for the mixture is:
ΔTf2 = Kf × m2 = 1.86 × 0.2 = 0.372 C
The freezing point of the mixture is:
Tf2 = Tf1 - ΔTf2 = -1.86 - 0.372 = -1.488 C
Therefore, the freezing point of the mixture is -1.488C.
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The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer?
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The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer?.
The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The freezing point of 0.1 M solution of glucose is -1.860C. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will bea)–7.44ºCb)–5.58ºCc)–3.72ºCd)–2.79ºCCorrect answer is option 'C'. Can you explain this answer?.
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