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The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
(Kf for benzene = 5.12 K kg mol–1)
(Kf for benzene = 5.12 K kg mol–1)
- a)64.6%
- b)80.4%
- c)74.6%
- d)94.6%
Correct answer is option 'D'. Can you explain this answer?
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The freezing point of benzene decreases by 0.45ºC when 0.2 g of a...
°C for every 1 mole of solute dissolved in 1000 grams of benzene. This is known as the cryoscopic constant (Kf) of benzene and has a value of 5.12 °C/m.
Using this information, we can calculate the number of moles of solute dissolved in benzene based on the observed decrease in freezing point.
Let's say the freezing point of pure benzene is -5.5°C and the freezing point of a solution is -6.7°C. The change in freezing point is:
ΔTf = Tf,pure - Tf,solution
ΔTf = -5.5°C - (-6.7°C)
ΔTf = 1.2°C
We can now use the cryoscopic constant to calculate the molality (m) of the solution:
ΔTf = Kf * m
m = ΔTf / Kf
m = 1.2°C / 5.12°C/m
m = 0.2344 mol/kg
This means that for every 1000 grams of benzene, 0.2344 moles of solute are dissolved. To calculate the mass of solute, we need to know the molar mass (M) of the solute:
mass of solute = m * M * 1000
For example, if the solute is naphthalene (C10H8) with a molar mass of 128.17 g/mol, the mass of solute in 1000 grams of benzene would be:
mass of solute = 0.2344 mol/kg * 128.17 g/mol * 1000 g = 30.03 g
Therefore, 30.03 grams of naphthalene are dissolved in 1000 grams of benzene, resulting in a freezing point depression of 1.2°C.
Using this information, we can calculate the number of moles of solute dissolved in benzene based on the observed decrease in freezing point.
Let's say the freezing point of pure benzene is -5.5°C and the freezing point of a solution is -6.7°C. The change in freezing point is:
ΔTf = Tf,pure - Tf,solution
ΔTf = -5.5°C - (-6.7°C)
ΔTf = 1.2°C
We can now use the cryoscopic constant to calculate the molality (m) of the solution:
ΔTf = Kf * m
m = ΔTf / Kf
m = 1.2°C / 5.12°C/m
m = 0.2344 mol/kg
This means that for every 1000 grams of benzene, 0.2344 moles of solute are dissolved. To calculate the mass of solute, we need to know the molar mass (M) of the solute:
mass of solute = m * M * 1000
For example, if the solute is naphthalene (C10H8) with a molar mass of 128.17 g/mol, the mass of solute in 1000 grams of benzene would be:
mass of solute = 0.2344 mol/kg * 128.17 g/mol * 1000 g = 30.03 g
Therefore, 30.03 grams of naphthalene are dissolved in 1000 grams of benzene, resulting in a freezing point depression of 1.2°C.
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The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer?
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The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. 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 freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. 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 freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer?.
The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. 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 freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. 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 freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer?.
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The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be(Kf for benzene = 5.12 K kg mol–1)a)64.6%b)80.4%c)74.6%d)94.6%Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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