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The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kf is 1.86° mol–1 kg, the depression in freezing point ΔTf is ____ °C.
(Round off up to 2 decimal places)
(Round off up to 2 decimal places)
Correct answer is '7.78'. Can you explain this answer?
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The mole fraction of a non-electrolyte in aqueous solution is 0.07. If...
- **Given Data**
The mole fraction of a non-electrolyte in aqueous solution is 0.07 and the Kf value is 1.86 °C mol⁻¹kg.
- **Calculation**
The formula to calculate the depression in freezing point (ΔTf) is:
ΔTf = Kf * m
Where:
ΔTf = depression in freezing point
Kf = cryoscopic constant (1.86 °C mol⁻¹kg)
m = molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since the solution is non-electrolyte, the molality is equal to the mole fraction of the solute.
- **Substitute Values**
m = 0.07
Kf = 1.86 °C mol⁻¹kg
- **Calculate ΔTf**
ΔTf = 1.86 * 0.07
ΔTf = 0.1302 °C
- **Final Answer**
Therefore, the depression in freezing point (ΔTf) is 0.1302 °C, which rounded off to 2 decimal places is 0.13 °C.
The mole fraction of a non-electrolyte in aqueous solution is 0.07 and the Kf value is 1.86 °C mol⁻¹kg.
- **Calculation**
The formula to calculate the depression in freezing point (ΔTf) is:
ΔTf = Kf * m
Where:
ΔTf = depression in freezing point
Kf = cryoscopic constant (1.86 °C mol⁻¹kg)
m = molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Since the solution is non-electrolyte, the molality is equal to the mole fraction of the solute.
- **Substitute Values**
m = 0.07
Kf = 1.86 °C mol⁻¹kg
- **Calculate ΔTf**
ΔTf = 1.86 * 0.07
ΔTf = 0.1302 °C
- **Final Answer**
Therefore, the depression in freezing point (ΔTf) is 0.1302 °C, which rounded off to 2 decimal places is 0.13 °C.
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Community Answer
The mole fraction of a non-electrolyte in aqueous solution is 0.07. If...
Given, mole fraction of non electrolyte = 0.07
Mole fraction of solvent = 1- Mole fraction of non electrolyte
Mole fraction of solvent = 1- 0.07 = 0.93
Kf = 1.86
ΔTf = Kf m
Mole fraction of solvent = 1- Mole fraction of non electrolyte
Mole fraction of solvent = 1- 0.07 = 0.93
Kf = 1.86
ΔTf = Kf m
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The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer?
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The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer?.
The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kfis 1.86° mol–1kg, the depression in freezing pointΔTfis ____ °C.(Round off up to 2 decimal places)Correct answer is '7.78'. Can you explain this answer?.
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