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1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.
- a)5 × 10–19
- b)3 × 10–19
- c)1 × 10–19
- d)7 × 10–19
Correct answer is option 'A'. Can you explain this answer?
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1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and...
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1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and...
To solve this problem, we need to use the concept of adsorption and the formula for surface area.
Given:
Mass of charcoal (m) = 1 g
Volume of acetic acid (V) = 100 mL = 0.1 L
Initial molarity of acetic acid (M1) = 0.5 M
Final molarity of acetic acid (M2) = 0.49 M
Surface area of charcoal (A) = 3.01 x 10^2 m^2/g
The amount of acetic acid adsorbed by the charcoal can be calculated using the equation:
n = (M1 - M2) x V
where n is the number of moles of acetic acid adsorbed.
Substituting the given values:
n = (0.5 - 0.49) x 0.1
n = 0.01 x 0.1
n = 0.001 mol
Now, we can calculate the number of molecules of acetic acid adsorbed:
Number of molecules (N) = n x Avogadro's number
where Avogadro's number is 6.022 x 10^23 mol^-1
N = 0.001 x 6.022 x 10^23
N = 6.022 x 10^20 molecules
To find the surface area adsorbed by each molecule of acetic acid, we divide the total surface area of charcoal by the number of molecules:
Surface area per molecule = A/N
Substituting the given values:
Surface area per molecule = (3.01 x 10^2)/(6.022 x 10^20)
Surface area per molecule = 5 x 10^-19 m^2
Therefore, the surface area of charcoal adsorbed by each molecule of acetic acid is 5 x 10^19 m^2.
Hence, the correct answer is option A: 5 x 10^19.
Given:
Mass of charcoal (m) = 1 g
Volume of acetic acid (V) = 100 mL = 0.1 L
Initial molarity of acetic acid (M1) = 0.5 M
Final molarity of acetic acid (M2) = 0.49 M
Surface area of charcoal (A) = 3.01 x 10^2 m^2/g
The amount of acetic acid adsorbed by the charcoal can be calculated using the equation:
n = (M1 - M2) x V
where n is the number of moles of acetic acid adsorbed.
Substituting the given values:
n = (0.5 - 0.49) x 0.1
n = 0.01 x 0.1
n = 0.001 mol
Now, we can calculate the number of molecules of acetic acid adsorbed:
Number of molecules (N) = n x Avogadro's number
where Avogadro's number is 6.022 x 10^23 mol^-1
N = 0.001 x 6.022 x 10^23
N = 6.022 x 10^20 molecules
To find the surface area adsorbed by each molecule of acetic acid, we divide the total surface area of charcoal by the number of molecules:
Surface area per molecule = A/N
Substituting the given values:
Surface area per molecule = (3.01 x 10^2)/(6.022 x 10^20)
Surface area per molecule = 5 x 10^-19 m^2
Therefore, the surface area of charcoal adsorbed by each molecule of acetic acid is 5 x 10^19 m^2.
Hence, the correct answer is option A: 5 x 10^19.
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1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer?
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1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer?.
1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer?.
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ample number of questions to practice 1 g of charcoal adsorbs 100 mL of 0.5 MCH3COOH to form a monolayer and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g.a)5 × 10–19b)3 × 10–19c)1 × 10–19d)7 × 10–19Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemistry tests.
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