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1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:
- a)3.015 × 102 m2
- b)5 × 10–19 m2
- c)2.5 × 10–19 m2
- d)1.25 × 10–19 m2
Correct answer is option 'B'. Can you explain this answer?
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1g charcoal show chemical adsorption of some acetic acid on coming in ...
We know that the volume of solution will remain unchanged after adsorption ( charcoal would presumably adsorb only the solute particles).
So, the no. of moles of AcH adsorbed = n = n1 -n2
where n1= initial moles = 0.5 M × 100ml = 0.5M×0.1 L
= 0.05 mol
n2= final moles = 0.49 M × 0.1L = 0.049 mol
Therefore, n= (0.05- 0.049 ) mol = 0.001 mol
No. of molecules of AcH adsorbed = n× Avogadro no.
i.e. N = 0.001 × 6.022 × 10^23 = 6.022 × 10^ 20
We know,
Surface area of adsorbent = Area of each molecule × N
So, area of each molecule = (3.015 × 10^2) ÷ (6.022 × 10^ 20)
*AREA OF EACH MOLECULE= 5.0 × 10 ^ -19*
So, the no. of moles of AcH adsorbed = n = n1 -n2
where n1= initial moles = 0.5 M × 100ml = 0.5M×0.1 L
= 0.05 mol
n2= final moles = 0.49 M × 0.1L = 0.049 mol
Therefore, n= (0.05- 0.049 ) mol = 0.001 mol
No. of molecules of AcH adsorbed = n× Avogadro no.
i.e. N = 0.001 × 6.022 × 10^23 = 6.022 × 10^ 20
We know,
Surface area of adsorbent = Area of each molecule × N
So, area of each molecule = (3.015 × 10^2) ÷ (6.022 × 10^ 20)
*AREA OF EACH MOLECULE= 5.0 × 10 ^ -19*
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1g charcoal show chemical adsorption of some acetic acid on coming in ...
To calculate the amount of acetic acid adsorbed by the charcoal, we can use the equation:
Adsorbed amount = initial concentration - final concentration
Given:
Initial concentration of acetic acid = 0.50 M
Final concentration of acetic acid = 0.49 M
Adsorbed amount = 0.50 M - 0.49 M
Adsorbed amount = 0.01 M
Now, to calculate the mass of acetic acid adsorbed, we need to know the volume of acetic acid used. In this case, it is given as 100 mL.
Using the equation:
Mass = molarity x volume x molar mass
Molarity of acetic acid = 0.01 M
Volume of acetic acid = 100 mL = 0.1 L (converted to liters)
Molar mass of acetic acid = 60.052 g/mol
Mass = 0.01 M x 0.1 L x 60.052 g/mol
Mass = 0.06 g
Therefore, the mass of acetic acid adsorbed by the charcoal is 0.06 grams.
Adsorbed amount = initial concentration - final concentration
Given:
Initial concentration of acetic acid = 0.50 M
Final concentration of acetic acid = 0.49 M
Adsorbed amount = 0.50 M - 0.49 M
Adsorbed amount = 0.01 M
Now, to calculate the mass of acetic acid adsorbed, we need to know the volume of acetic acid used. In this case, it is given as 100 mL.
Using the equation:
Mass = molarity x volume x molar mass
Molarity of acetic acid = 0.01 M
Volume of acetic acid = 100 mL = 0.1 L (converted to liters)
Molar mass of acetic acid = 60.052 g/mol
Mass = 0.01 M x 0.1 L x 60.052 g/mol
Mass = 0.06 g
Therefore, the mass of acetic acid adsorbed by the charcoal is 0.06 grams.
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1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer?
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1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. 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 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer?.
1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. 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 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer?.
Solutions for 1g charcoal show chemical adsorption of some acetic acid on coming in contact with 100 mL of 0.50 M acetic acid and the molarity of acetic acid is reduced to 0.49 M. If surface area of charcoal is 3.015 × 102 m2/g, the surface area of each chemical adsorbed molecule of acetic acid is:a)3.015 × 102 m2b)5 × 10–19 m2c)2.5 × 10–19 m2d)1.25 × 10–19 m2Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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