1 g of activated charcoal has a surface area 105 m2. If there is complete monolayer coverage and has effective surface area of NH3 is 0.015 nm2, then volume of NH3 in cm3 at STP that could be adsorbed on 50 g of charcoal is_______.

Total surface area available for adsorption =50×103m2 Effective surface area of one NH3 molecule =1.5×10−20m2∴ No. of molecules of NH3 adsorbed =50×1031.5×10−20=33.3×1017 Moles of NH3=33.3×10176×1023=5.5×10−6∴VNH3=nRTP=5.5×10−6×0.082×2731=1.23×10−4 litre =0.123cm3