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one gram of activated carbon has a surfa...

one gram of activated carbon has a surface are of `1000m^(2)`. Considering complete coverage as well as monomolecular adsorption , how much ammonia at 1 atm and 273 K would be absorbed on the surface of `44/7` g carbon if radius of a ammonia molecules is `10^(-8) cm`.

A

7.46L

B

0.33L

C

44.8L

D

23.5L

Text Solution

Verified by Experts

The correct Answer is:
A
Total surface area of carbon `= 4xx10^(7) cm^(2)`?
`r=10^(-8) cm`
Surface area of `NH_(3) = pir^(2)`
`=(22)/(7)xx10^(-16)cm^(2)`
No. of `NH_(3)` molecules adsorbed =`=((44)/(7)xx10^(7))/((22)/(7)xx10^(-16))`
Vol. of `NH_(3)` adsorbed at 1 atm 273 K
`=(2xx10^(23))/(6xx10^(23))xx22.4=77.46L`
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