Question

One gram of activated carbon has a surface area of $1000m^2$ . Considering complete coverage as well as monomolecular adsorption, how much ammonia at $1$ atm and $273$ K would be adsorbed on the surface of $\frac{44}{7}$ g carbon if radius of ammonia molecules is ${10}^{-8}$cm.

Answer 1

Total surface area of carbon $=\frac{44}{7}\times {10}^{7}c{m}^2$
$r={10}^{-8}cm$
Surface area of $N{H}_3=\pi r^2$
$=\frac{22}{7}\times {10}^{-16}c{m}^2$
No. of $N{H}_3$ molecules adsorbed $=\frac{\frac{44}{7}\times {10}^7}{\frac{22}{7}\times {10}^{-16}}$
$=2\times {10}^{23}$
Vol. of $N{H}_3$ adsorbed at STP $=\frac{2\times {10}^{23}}{6\times {10}^{23}}\times 22.4$
$=7.46L$