# Example-1

Example-1

One gram of activates carbon has a surface area of 1000 m

^{2}. Considering complete coverage as well as monomolecular adsorption, how much ammonia at STP would be adsorbed on the surface of 44/7 g carbon if the radius of ammonia molecules is 10^{â€“8}cm. [*N*= 6 Ã—10_{A}^{23}]- 7.46 L
- 0.33 L
- 44.8 L
- 23.5 L

Solution(A)

Total surface area of carbon =

*r*= 10

^{â€“8}cm

Surface area of NH

_{3}=*Ï€**r*^{2}=Number of NH

_{3}molecules adsorbed = = 2 Ã— 10^{23}Volume of NH

_{3}adsorbed at STP = = 7.46 L# Example-2

Example-2

At STP the volume of nitrogen gas required to cover a sample of silica gel, assuming Langmuir monolayer adsorption, is found to be 1.30 cm

^{3}/g of the gel. The area occupied by a nitrogen molecule is 0.16 nm^{2}. What is the surface area per gram of silica gel? (Given*N*= 6 Ã— 10_{A}^{23})- 5.568 m
^{2}/g - 3.48 m
^{2}/g - 1.6 m
^{2}/g - None of these

Solution (A)

Number of molecules per gram of N

_{2}in monolayer == 3.48 Ã— 10

^{19}Cross-sectional area of a molecule = 1.6 Ã— 10

^{â€“19}m^{2}âˆ´ Area covered by molecules per gram = 3.48 Ã— 10

^{19}Ã— 1.6 Ã— 10^{â€“19}= 5.568 m^{2}âˆ´ Surface area = 5.568 m

^{2}/g# Example-3

Example-3

About 10% sites of catalyst bed have absorbed by H

_{2}. On heating, H_{2}gas is evolved from sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm^{3}. Number of sites available is 6.0 Ã— 10^{15}per cm^{2}and surface area is 1000 cm^{2}. Find out the number of surface sites occupied per molecule of H_{2}. (Given*N*= 6 Ã— 10_{A}^{23})- 1
- 2
- 3
- none of these

Solution (C)

Adsorbed moles of H

_{2}= = 3 Ã— 10^{â€“6}âˆ´ Number of absorbed molecules of H

_{2}= 3 Ã— 10^{â€“6}Ã— 6 Ã— 10^{23}=18 Ã— 10^{17}Total number of surface sites available = 6 Ã— 10

^{5}Ã— 1000 = 6 Ã— 10^{18}cm^{2}Number of surface sites that is occupied by adsorption of H

_{2}= = 6 Ã— 10

^{17}Number of surface sites occupied by one molecule of H

_{2}=# Example-4

Example-4

A sample of 16 g charcoal was brought into contact with a CH

_{4}gas contained in a vessel of 1 L at 27Â°C. The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6 g/cm^{3}. What is the volume of the CH_{4}gas adsorbed per gram of the adsorbent at 608 torr and 27Â°C?- 125 mL/g
- 16.25 mL/g
- 26 mL/g
- None of these

Solution (B)

Final volume of gas at 608 torr pressure is given by

*V*

_{2}= = = 1.25

or

*V*_{2}= 1250 mLVolume occupied by gas = Volume of vessel â€“ Volume occupied by charcoal

= 1000 â€“ mL

Different of volume is due to adsorption of gas by charcoal. Therefore,

Volume of gas adsorbed by chemical = 1250 â€“ 990 = 260 mL

Volume of the gas adsorbed per gram of charcoal

= at 208 torr and 27Â°C.

# Example-5

Example-5

The density of gold is . If of gold is dispersed in 1 L of water to give a sol having spherical gold particles of radius 10 nm, then the number of gold particles per of the sol will be:

Solution (D)

Volume of the gold dispersed in 1 L water

Volume of the gold sol particle

Number of gold sol particle in

Number of gold sol particle in 1