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2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?
- a)80%
- b)99%
- c)75%
- d)100%
Correct answer is option 'B'. Can you explain this answer?
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2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depressi...
To find the percentage association of the acid, we need to calculate the degree of dissociation. The degree of dissociation (α) is the fraction of the acid molecules that dissociate into ions in the solution.
Given data:
- Mass of benzoic acid (m1) = 2.0 g
- Mass of benzene (m2) = 25.0 g
- Depression in freezing point (ΔTf) = 1.62 K
- Molal depression constant of benzene (kf) = 4.9 K kg mol⁻¹
First, let's calculate the molality of the solution using the formula:
molality (m) = moles of solute / mass of solvent in kg
We know the molar mass of benzoic acid is 122.12 g/mol. So, the moles of benzoic acid (n1) can be calculated as:
n1 = m1 / molar mass of benzoic acid
= 2.0 g / 122.12 g/mol
= 0.01636 mol
The mass of the solvent benzene (m2) is given as 25.0 g. Converting it to kg:
mass of benzene (m2) = 25.0 g = 0.025 kg
Now, we can calculate the molality (m):
m = n1 / m2
= 0.01636 mol / 0.025 kg
= 0.6544 mol/kg
Next, we can use the formula for the freezing point depression:
ΔTf = kf * m * i
Where i is the van't Hoff factor, which represents the number of particles into which the solute dissociates. For benzoic acid, it is 1, as it does not dissociate into ions.
Substituting the given values:
1.62 K = 4.9 K kg mol⁻¹ * 0.6544 mol/kg * 1
1.62 K = 3.20256 K
Now, let's calculate the degree of dissociation (α):
α = ΔTf / kf * m
= 1.62 K / 3.20256 K
= 0.5059
The degree of dissociation of benzoic acid is approximately 0.5059 or 50.59%.
As the question asks for the percentage association, we subtract the degree of dissociation from 1:
Percentage association = 1 - 0.5059
= 0.4941
= 49.41%
Therefore, the closest approximation for the percentage association of the acid is 49% or option C.
Given data:
- Mass of benzoic acid (m1) = 2.0 g
- Mass of benzene (m2) = 25.0 g
- Depression in freezing point (ΔTf) = 1.62 K
- Molal depression constant of benzene (kf) = 4.9 K kg mol⁻¹
First, let's calculate the molality of the solution using the formula:
molality (m) = moles of solute / mass of solvent in kg
We know the molar mass of benzoic acid is 122.12 g/mol. So, the moles of benzoic acid (n1) can be calculated as:
n1 = m1 / molar mass of benzoic acid
= 2.0 g / 122.12 g/mol
= 0.01636 mol
The mass of the solvent benzene (m2) is given as 25.0 g. Converting it to kg:
mass of benzene (m2) = 25.0 g = 0.025 kg
Now, we can calculate the molality (m):
m = n1 / m2
= 0.01636 mol / 0.025 kg
= 0.6544 mol/kg
Next, we can use the formula for the freezing point depression:
ΔTf = kf * m * i
Where i is the van't Hoff factor, which represents the number of particles into which the solute dissociates. For benzoic acid, it is 1, as it does not dissociate into ions.
Substituting the given values:
1.62 K = 4.9 K kg mol⁻¹ * 0.6544 mol/kg * 1
1.62 K = 3.20256 K
Now, let's calculate the degree of dissociation (α):
α = ΔTf / kf * m
= 1.62 K / 3.20256 K
= 0.5059
The degree of dissociation of benzoic acid is approximately 0.5059 or 50.59%.
As the question asks for the percentage association, we subtract the degree of dissociation from 1:
Percentage association = 1 - 0.5059
= 0.4941
= 49.41%
Therefore, the closest approximation for the percentage association of the acid is 49% or option C.
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Community Answer
2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depressi...
Normal molecular weight of C6H5COOH
= 72 + 5 + 12 + 32 + 1 = 122
% association = 0.992 x 100
= 99.2 %
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2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. Can you explain this answer?
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2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. 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 2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. Can you explain this answer?.
2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. 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 2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.0 g of benzoic acid dissolved in 25.0 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant kf of benzene is 4.9 K kg mol-1. What is the percentage association (closest approximation) of the acid?a)80%b)99%c)75%d)100%Correct answer is option 'B'. Can you explain this answer?.
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