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1 g of activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.
Correct answer is between '0.322,0.323'. Can you explain this answer?
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1 g of activated charcoal has a surface area 103 m3. If complete monol...
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1 g of activated charcoal has a surface area 103 m3. If complete monol...
**Given Data:**
- Surface area of activated charcoal = 103 m^2/g
- Effective surface area of NH3 molecule = 0.129 cm^2
- Mass of charcoal = 25 g
**Calculating the Number of Moles of Charcoal:**
1. Calculate the number of moles of charcoal using the formula:
Number of moles = Mass / Molar mass
- Molar mass of charcoal = 12.01 g/mol (approximate value of carbon)
- Mass of charcoal = 25 g
Number of moles of charcoal = 25 g / 12.01 g/mol ≈ 2.08 mol
**Calculating the Number of Charcoal Particles:**
2. Calculate the number of charcoal particles using Avogadro's number:
Number of particles = Number of moles x Avogadro's number
- Avogadro's number = 6.022 x 10^23 particles/mol
- Number of moles of charcoal = 2.08 mol
Number of charcoal particles = 2.08 mol x 6.022 x 10^23 particles/mol ≈ 1.25 x 10^24 particles
**Calculating the Number of NH3 Molecules Adsorbed:**
3. Calculate the number of NH3 molecules adsorbed assuming complete monolayer coverage:
Number of NH3 molecules = Number of particles x Surface area of 1 molecule / Effective surface area of NH3 molecule
- Number of particles = 1.25 x 10^24 particles
- Surface area of 1 molecule = 1 x 10^-4 cm^2 (approximate value)
- Effective surface area of NH3 molecule = 0.129 cm^2
Number of NH3 molecules = (1.25 x 10^24 particles) x (1 x 10^-4 cm^2) / (0.129 cm^2) ≈ 9.69 x 10^23 molecules
**Converting the Number of NH3 Molecules to Volume:**
4. Convert the number of NH3 molecules to volume using Avogadro's law:
Volume of NH3 = Number of molecules x Molar volume
- Molar volume at STP = 22.4 L/mol
- Number of NH3 molecules = 9.69 x 10^23 molecules
Volume of NH3 = (9.69 x 10^23 molecules) x (22.4 L/mol) / (6.022 x 10^23 molecules/mol) ≈ 0.322 L ≈ 322 cm^3
Therefore, approximately 0.322 cm^3 of NH3 at STP could be adsorbed on 25 g of charcoal.
- Surface area of activated charcoal = 103 m^2/g
- Effective surface area of NH3 molecule = 0.129 cm^2
- Mass of charcoal = 25 g
**Calculating the Number of Moles of Charcoal:**
1. Calculate the number of moles of charcoal using the formula:
Number of moles = Mass / Molar mass
- Molar mass of charcoal = 12.01 g/mol (approximate value of carbon)
- Mass of charcoal = 25 g
Number of moles of charcoal = 25 g / 12.01 g/mol ≈ 2.08 mol
**Calculating the Number of Charcoal Particles:**
2. Calculate the number of charcoal particles using Avogadro's number:
Number of particles = Number of moles x Avogadro's number
- Avogadro's number = 6.022 x 10^23 particles/mol
- Number of moles of charcoal = 2.08 mol
Number of charcoal particles = 2.08 mol x 6.022 x 10^23 particles/mol ≈ 1.25 x 10^24 particles
**Calculating the Number of NH3 Molecules Adsorbed:**
3. Calculate the number of NH3 molecules adsorbed assuming complete monolayer coverage:
Number of NH3 molecules = Number of particles x Surface area of 1 molecule / Effective surface area of NH3 molecule
- Number of particles = 1.25 x 10^24 particles
- Surface area of 1 molecule = 1 x 10^-4 cm^2 (approximate value)
- Effective surface area of NH3 molecule = 0.129 cm^2
Number of NH3 molecules = (1.25 x 10^24 particles) x (1 x 10^-4 cm^2) / (0.129 cm^2) ≈ 9.69 x 10^23 molecules
**Converting the Number of NH3 Molecules to Volume:**
4. Convert the number of NH3 molecules to volume using Avogadro's law:
Volume of NH3 = Number of molecules x Molar volume
- Molar volume at STP = 22.4 L/mol
- Number of NH3 molecules = 9.69 x 10^23 molecules
Volume of NH3 = (9.69 x 10^23 molecules) x (22.4 L/mol) / (6.022 x 10^23 molecules/mol) ≈ 0.322 L ≈ 322 cm^3
Therefore, approximately 0.322 cm^3 of NH3 at STP could be adsorbed on 25 g of charcoal.
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1 g of activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. Can you explain this answer?
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1 g of activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. 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 activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. 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 activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. Can you explain this answer?.
1 g of activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. 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 activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. 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 activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. Can you explain this answer?.
Solutions for 1 g of activated charcoal has a surface area 103 m3. If complete monolayer coverage is assumed and effective surface area of NH3 molecule is 0.129 cm2, how much NH3 in cm3 at STP could be adsorbed on 25 g of charcoal.Correct answer is between '0.322,0.323'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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