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 a diameter of $N{{H}_{3}}$ molecule is 0.3 nm.

Answer 1

Activated charcoal is the charcoal which is processed to have a large number of pores on its surface to accommodate a large number of solute particles on its surface through a process called adsorption.

Complete step by step answer:

- In the question it is asked how much ammonia is adsorbed on the surface of activated charcoal.

- In the question it is given that the surface area of the activated charcoal is ${{10}^{3}}{{m}^{2}}$ .

- The amount of charcoal given in the question is 25 g = (25) (1000) = 25000 ${{m}^{2}}$

- The diameter of the ammonia molecule is 0.3 nm.

- So, the radius becomes 0.15

- Surface area of ammonia molecule =$\pi \times r^{2}$

= $\pi \times 0.15nm \times 0.15nm$

= 0.07nm²

=$.0.7 \times 10^{-19}m^{2}$

- There is formula to calculate number of ammonia molecules = $\dfrac{25000 {{m}^{2}}}{0.710^{-19}m²}

- The number of moles of ammonia $=3.57 \times 10^{23}$

- Total total no. of ammonia molecules/Avogadro number

= $\dfrac{=3.57 \times 10^{23}}{6.023 \times 10^{23}}$

= 0.59

- We got the number of moles of ammonia adsorbed on the surface of activated charcoal. From this we can get the volume of ammonia as follows.

- Volume of ammonia adsorbed = 0.5922.4= 13.2 litres

- Therefore the volume of ammonia adsorbed is 13.2 litres

So, the correct answer is 13.2 litres

Note: Generally activated metals surfaces have a large number of pores. In this pores the gas particles are going to be accommodated through the process of adsorption. Adsorption is a physical phenomenon and it is a surface property.