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?

Question

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?

Answer

**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.

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