One gram of activated charcoal has a surface area of 1000m². If complete coverage is assumed, Approximately  how much ammonia (in cm³ at NTP) could be adsorbed on the surface of 25g of the charcoal ? (Given : Diameter of NH₃ molecule = 0.3nm)

Total surface area available = $=(25\times {10}^3)m^2=2.5\times {10}^{8}c{m}^2$
Surface area covered by one molecule of  $N{H}_3=\pi r^2$
$=[\frac{22}{7}\times {(\frac{0.3}{2}\times {10}^{-7})}^2]c{m}^2=(\frac{22}{7}\times 0.15\times {10}^{-14}\times 0.15)c{m}^2$$=0.0707\times {10}^{14}c{m}^2=7.07\times {10}^{-16}c{m}^2$

No. of ammonia molecules  $=\frac{2.5\times {10}^8}{7.07\times {10}^{-16}}=3.536\times {10}^{23}$
No. of moles of  $N{H}_3=\frac{3.536\times {10}^{23}}{6.023\times {10}^{23}}=0.587$
Volume of $N{H}_3$  at NTP $=(0.587\times 22400)=$ =13.282.4cm³