Answer:
a) mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg
b) all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years
Explanation:
first we calculate the moles of carbon
moles = mass/molar mass
= 1kg/12gmol⁻¹
= 1000g/12gmol⁻¹
= 83.33 mol
now using the ideal gas equation
we find the volume of co₂required based on 83.33 moles
PVco₂ = nRT
Vco₂ = nRT/P
Vco₂ = (83.22mol × 0.0821L atm k⁻¹ mol⁻¹ 298 K) / 1 atm
Vco₂ = 2083.73 L
so since CO₂ in air is 0.0390% by volume in the atmosphere, we find the the total amount of air required to obtain 1kg carbon
therefore
Vair × 0.0390/100 = 2038.73L
Vair = (2038.73L × 100) / 0.0390
Vair = 5.23 × 10⁶L
therefore 5.23 × 10⁶ L of air will be required to obtain 1kg carbon
a)
Here we calculate the mass of air over 1 square meter of surface.
Remember that atmospheric pressure is the consequence of the force exerted by all the air above the surface; 1 bar is equivalent to 1.020×10⁴kgm⁻²
NOW
mass of air = 1.020×10⁴kgm⁻² × 1m²
= 1.020×10⁴kg
= 1.020×10⁷g [1kg = 10³g]
we now find the moles of air associated with it
moles = mass/molar mass
= 1.020 × 10⁷g / ( 20%×Mo₂ + 80%×Mn₂)
= 1.020 × 10⁷g / ( 20%×32gmol⁻¹ + 80%×28gmol⁻¹)
= 1.020 × 10⁷g / 28.8 gmol⁻¹
= 354166.67mol
so based on the question, for each mole (air), there is 0.0390% of CO₂
now to calculate the moles of CO₂ we say;
MolesCo₂ = 0.0390/100 × 354166.67mol
= 138.125 moles
Now we calculate mass of CO₂ from the above findings
Moles = mass/molar mass
mass = moles × molar mass
= 138.125 moles × 12gmol⁻¹
= 1657.5g
we covert to KG
= 1657.5g / 1000
mass = 1.65kg
therfore mass of carbon directly above 1 ( each) square meter of the earth is 1.65kg
b)
to find the number years required to use up all the CO₂, WE SAY
Number of years = total carbon per m² of the forest / carbon used up per m² from the forest per year
Number of years = 1.65kgm⁻² / 1kg²year⁻¹
Number of years = 1.65 years
Therefore all CO₂ will definitely be used up from the atmosphere directly above a forest in 1.65 years
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, where 
(in some books 6.022).