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Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.
Molecule of acetic acids adsorbed —
- a)6.023 x 1025
- b)6.023 x 1021
- c)6.023 x 1022
- d)6.023 x 1023
Correct answer is option 'A'. Can you explain this answer?
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Adsorption is the presence of excess concentration of any particular c...
Adsorption of hydrogen gas
- 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal.
- Density of H2 is 0.07gm/cc.
- Consider only monolayer adsorption.
Adsorption of acetic acid
- 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer.
- Polarity of CH3COOH reduces to 0.49.
- Need to calculate the number of molecules of acetic acid adsorbed.
Calculation
- First, calculate the surface area of the charcoal using the amount of hydrogen adsorbed:
- Volume of 1 gram of charcoal = 1/0.07 = 14.29 cm3
- Volume of H2 adsorbed per gram of charcoal = 1.12 cm3
- So, volume of H2 adsorbed per cm2 of charcoal surface = 1.12/14.29 = 0.0784 cm3/cm2
- Assuming a monolayer adsorption, the surface area of the charcoal can be calculated as follows:
- Area of 1 H2 molecule = 2 x (4/3) x π x (52 x 10-8)2 = 1.256 x 10-15 cm2
- Number of H2 molecules adsorbed per cm2 of charcoal surface = 0.0784/1.256 x 10-15 = 6.231 x 1013
- So, the surface area of the charcoal = number of H2 molecules adsorbed/area per H2 molecule = 6.231 x 1013/1.256 x 10-15 = 4.95 x 1028 cm2
- Next, calculate the number of acetic acid molecules adsorbed:
- The change in polarity of acetic acid indicates that one of the two CH3 groups is adsorbed on the charcoal surface.
- So, the number of acetic acid molecules adsorbed = number of CH3 groups adsorbed = (mass of charcoal used x number of CH3 groups per molecule)/(molar mass of CH3 group)
- Mass of charcoal used = 1 g
- Number of CH3 groups per molecule of acetic acid = 2
- Molar mass of CH3 group = 15 g/mol
- Therefore, number of acetic acid molecules adsorbed = (1 x 2)/(15 x 10-3) = 1.33 x 1022
- Since each mole of a substance contains 6.023 x 1023 molecules, the number of acetic acid molecules adsorbed = 6.023 x 1025 x 1.33 x 1022/1 mol = 6.023 x 1025/75 = 8.03 x 1022
- However, the given options do not match the calculated answer. It is possible that a mistake was made in the calculation or that the options are incorrect.
- 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal.
- Density of H2 is 0.07gm/cc.
- Consider only monolayer adsorption.
Adsorption of acetic acid
- 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer.
- Polarity of CH3COOH reduces to 0.49.
- Need to calculate the number of molecules of acetic acid adsorbed.
Calculation
- First, calculate the surface area of the charcoal using the amount of hydrogen adsorbed:
- Volume of 1 gram of charcoal = 1/0.07 = 14.29 cm3
- Volume of H2 adsorbed per gram of charcoal = 1.12 cm3
- So, volume of H2 adsorbed per cm2 of charcoal surface = 1.12/14.29 = 0.0784 cm3/cm2
- Assuming a monolayer adsorption, the surface area of the charcoal can be calculated as follows:
- Area of 1 H2 molecule = 2 x (4/3) x π x (52 x 10-8)2 = 1.256 x 10-15 cm2
- Number of H2 molecules adsorbed per cm2 of charcoal surface = 0.0784/1.256 x 10-15 = 6.231 x 1013
- So, the surface area of the charcoal = number of H2 molecules adsorbed/area per H2 molecule = 6.231 x 1013/1.256 x 10-15 = 4.95 x 1028 cm2
- Next, calculate the number of acetic acid molecules adsorbed:
- The change in polarity of acetic acid indicates that one of the two CH3 groups is adsorbed on the charcoal surface.
- So, the number of acetic acid molecules adsorbed = number of CH3 groups adsorbed = (mass of charcoal used x number of CH3 groups per molecule)/(molar mass of CH3 group)
- Mass of charcoal used = 1 g
- Number of CH3 groups per molecule of acetic acid = 2
- Molar mass of CH3 group = 15 g/mol
- Therefore, number of acetic acid molecules adsorbed = (1 x 2)/(15 x 10-3) = 1.33 x 1022
- Since each mole of a substance contains 6.023 x 1023 molecules, the number of acetic acid molecules adsorbed = 6.023 x 1025 x 1.33 x 1022/1 mol = 6.023 x 1025/75 = 8.03 x 1022
- However, the given options do not match the calculated answer. It is possible that a mistake was made in the calculation or that the options are incorrect.
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Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer?
Question Description
Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. 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 Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer?.
Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. 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 Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer?.
Solutions for Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Adsorption is the presence of excess concentration of any particular component at the surface of liquid or solid phase as compared to bulk. This is due to presence of residual forces at the surface of body. In the adsorption of hydrogen gas over a sample, of charcoal, 1.12cm3 of H2(g) measured over S.T.P. was found to adsorb per gram of charcoal. Consider only monolayer adsorption. Density of H2 is 0.07gm/cc. In another experiment same 1 gm charcoal adsorbs 100m1 of 0.5M CH3COOH to form monolayer and thereby the polarity of CH3COOH reduces to 0.49.(3 x 47.73 )1/347t =2.24.(2.24)2 = 5Q.Molecule of acetic acids adsorbed a)6.023 x 1025b)6.023 x 1021c)6.023 x 1022d)6.023 x 1023Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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