Question

One gram of activated charcoal has a surface area of ${10}^3{m}^2$. If complete coverage by monolayer is assumed, how much $N{H}_3$ in litres at STP would be adsorbed on the surface of $25$ g of the charcoal? Given diameter of $N{H}_3$ molecule is $0.3$ nm. Round your answer to the nearest integer.

Answer 1

The surface area of 25 g of charcoal is $25\times 1000=25000m^2$
Surface area occupied by one ammonia molecule $=0.3\times 0.3=0.09n{m}^2=9\times {10}^{-20}{m}^2$
Total number of ammonia molecules $=\frac{25000}{9\times {10}^{-20}}=2.77\times {10}^{23}$
The number of moles of ammonia $=\frac{2.77\times {10}^{23}}{6.023\times {10}^{23}}=0.46$
Volume of ammonia adsorbed $22.4\times 0.46=10.3L$