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Example-1

Example-1
One gram of activates carbon has a surface area of 1000 m2. 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 10–8cm. [NA = 6 ×1023]
  1. 7.46 L
  2. 0.33 L
  3. 44.8 L
  4. 23.5 L
Solution(A)
Total surface area of carbon = Description: 38022.png
 
r = 10–8 cm
 
Surface area of NH3 = πr2 = Description: 38031.png
 
Number of NH3 molecules adsorbed = Description: 38040.png = 2 × 1023
 
Volume of NH3 adsorbed at STP = Description: 38052.png = 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 cm3/g of the gel. The area occupied by a nitrogen molecule is 0.16 nm2. What is the surface area per gram of silica gel? (Given NA = 6 × 1023)
  1. 5.568 m2/g
  2. 3.48 m2/g
  3. 1.6 m2/g
  4. None of these
Solution (A)
Number of molecules per gram of N2 in monolayer = Description: 38065.png
= 3.48 × 1019
 
Cross-sectional area of a molecule = 1.6 × 10–19 m2
 
∴ Area covered by molecules per gram = 3.48 × 1019 × 1.6 × 10–19 = 5.568 m2
 
∴ Surface area = 5.568 m2/g
 

Example-3

Example-3
About 10% sites of catalyst bed have absorbed by H2. On heating, H2 gas is evolved from sites and collected at 0.03 atm and 300 K in a small vessel of 2.46 cm3. Number of sites available is 6.0 × 1015 per cm2 and surface area is 1000 cm2. Find out the number of surface sites occupied per molecule of H2. (Given NA = 6 × 1023)
  1. 1
  2. 2
  3. 3
  4. none of these
Solution (C)
Adsorbed moles of H2 = Description: 38074.png = 3 × 10–6
∴ Number of absorbed molecules of H2 = 3 × 10–6 × 6 × 1023 =18 × 1017
 
Total number of surface sites available = 6 × 105 × 1000 = 6 × 1018 cm2
 
Number of surface sites that is occupied by adsorption of H2
 
Description: 38082.png = 6 × 1017
 
Number of surface sites occupied by one molecule of H2 = Description: 38091.png
 

Example-4

Example-4
A sample of 16 g charcoal was brought into contact with a CH4 gas contained in a vessel of 1 L at 27°C. The pressure of gas was found to fall from 760 to 608 torr. The density of charcoal sample is 1.6 g/cm3. What is the volume of the CH4 gas adsorbed per gram of the adsorbent at 608 torr and 27°C?
  1. 125 mL/g
  2. 16.25 mL/g
  3. 26 mL/g
  4. None of these
Solution (B)
Final volume of gas at 608 torr pressure is given by
 
V2 = Description: 38099.png = Description: 38106.png = 1.25
 
or V2 = 1250 mL
 
Volume occupied by gas = Volume of vessel – Volume occupied by charcoal
 
= 1000 – Description: 38114.png mL
 
Different of volume is due to adsorption of gas by charcoal. Therefore,
 
Volume of gas adsorbed by chemical = 1250 – 990 = 260 mL
 
Volume of the gas adsorbed per gram of charcoal
 
Description: 38122.png at 208 torr and 27°C.
 

Example-5

Example-5
The density of gold is Description: 38136.png. If Description: 38145.png 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 Description: 38155.png of the sol will be:
  1. Description: 38163.png
  2. Description: 38172.png
  3. Description: 38181.png
  4. Description: 38188.png
Solution (D)
Volume of the gold dispersed in 1 L water
Description: 38199.pngDescription: 38206.pngDescription: 38214.png
 
Radius of gold sol particle Description: 38222.pngDescription: 38232.pngDescription: 38241.png
 
Volume of the gold sol particle Description: 38252.pngDescription: 38260.pngDescription: 38276.png
 
Number of gold sol particle in Description: 38284.png Description: 38291.png Description: 38298.png
 
Number of gold sol particle in 1 Description: 38305.png Description: 38313.png
 




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