Home
Class 12
CHEMISTRY
one gram of activated carbon has a surfa...

one gram of activated carbon has a surface are of `1000m^(2)`. Considering complete coverage as well as monomolecular adsorption , how much ammonia at 1 atm and 273 K would be absorbed on the surface of `44/7` g carbon if radius of a ammonia molecules is `10^(-8) cm`.

A

7.46 L

B

0.33 L

C

44.8 L

D

23.5 L

Text Solution

Verified by Experts

Total surface area of carbon=`44/7`
`r=10^8 c`
Surface area of `NH_3+p^2=22/7-162`
Number of `NH_3` molecules adsorbed `(44/7 times0^7)/(22/7 times 0^(16)) 210^2,`
Vol. of `NH_3` adsorbed at STP =2.4
Promotional Banner

Topper's Solved these Questions

  • SURFACE CHEMISTRY

    VMC MODULES ENGLISH|Exercise JEE Main (Archive)|14 Videos

  • SURFACE CHEMISTRY

    VMC MODULES ENGLISH|Exercise JEE MAIN ARCHIVE|25 Videos

  • SURFACE CHEMISTRY

    VMC MODULES ENGLISH|Exercise LEVEL-1|74 Videos

  • STRUCTURE OF ATOM

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-F|7 Videos

  • THE SOLID STATE

    VMC MODULES ENGLISH|Exercise EXERCISE-J|10 Videos

Similar Questions

Explore conceptually related problems

The volume of nitrogen gas Um (measured at STP) required to cover a sample of silica get with a mono-molecular layer is 129 cm^(3) g^(-1) of gel. Calculate the surface area per gram of the gel if each nitrogen molecule occupies 16.2 xx 10^(-20) m^(2).

The volume of nitrogen gas Vm (at STP) reqired to cover a sample of silica gel with a monomolecular layer is 129cm^(3)g^(-1) of gel. Calculate the surface area per gram of the gel if each nitrogen molecule occupies 16.23xx10^(-20)m^(2) .

Finely divided catalyst has greater surface area and has greater catalytic activity than the compact solid. If a total surface area of 6291456 cm^(2) is required for adsorption in a catalytic gaseous reaction, then how many splits should be made in a cube of exactly 1 cm in length to achieve required surface area? [Given : One split of a cube gives eight cubes of same size]

A point source is emitting 0.2 W of ultravio- let radiation at a wavelength of lambda = 2537 Å . This source is placed at a distance of 1.0 m from the cathode of a photoelectric cell. The cathode is made of potassium (Work function = 2.22 eV ) and has a surface area of 4 cm^(2) . (a) According to classical theory, what time of exposure to the radiation shall be required for a potassium atom to accumulate sufficient energy to eject a photoelectron. Assume that radius of each potassium atom is 2 Å and it absorbs all energy incident on it. (b) Photon flux is defined as number of light photons reaching the cathode in unit time. Calculate the photon flux. (c) Photo efficiency is defined as probability of a photon being successful in knocking out an electron from the metal surface. Calculate the saturation photocurrent in the cell assuming a photo efficiency of 0.1. (d) Find the cut – off potential difference for the cell.

1 L of 0.6 M acetic acid is shaken with 2 g activated carbon.Activated carbon absorbs some acetic acid on its surface only.This process is called adsorption.The final concentration of the solution after adsorption is 0.5 M.What is the amount of acetic acid absorbed per gram of carbon.

One gram of charcoal adsorbs 400 " mL of " 0.5 M acetic acid to form a mono layer and the molarity of acetic acid reduced to 0.49. Calculate the surface area of charcoal adsorbed by each molecule of acetic acid. The surface area of charcoal is 3.01xx10^(2)m^(2)g^(-1) .

Volume of N_(2) at 1 atm, 273 K required to form a monolayer on the surface of iron catalyst is 8.15ml//gm of the adsorbent. What will be the surface area of the adsorbent per gram if each nitrogen molecule occupies 16 xx 10^(-22)m^(2) ? [Take : N_(A)=6xx10^(23) ]

If the water molecules in 1.0 g of water were distributed uniformly over the surface of earth, how many such molecules would there be in 1.0 cm^2 of earth's surface ?

Twenty precent of the surface sites of a catalyst is occupied by nitrogen molecules. The density of surface sites is 6.023xx10^(14)cm^(-2) . The total sarface area is 1000cm^(2) . The catalyst is is henced to 300K and nitrogen is completely desorbed a pressure of 0.001 atm and volume of 2.46cm^(3) . Calculate the number of sites occupied by niitrogen molecules.

At 1 atm and 273 K 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^(-1) of the gel. The area occupied by a nitrogen molecule is 0.16nm^(2) . Find out the no. of surface sites occupied per molecule of N_(2) .

VMC MODULES ENGLISH-SURFACE CHEMISTRY-LEVEL-2
  1. The coagulation of 200 mL of a positive colloid took place when 0.73 g...

    Text Solution

    |

  2. A detergent C2H2 solution becomes a colloidal sol at a concentration ...

    Text Solution

    |

  3. one gram of activated carbon has a surface are of 1000m^(2). Consideri...

    Text Solution

    |

  4. AT STP the volume of nitrogen gas required to cover a sample of silica...

    Text Solution

    |

  5. 10 % sites of catalyst bed have adsorbed by H(2) on Heating H(2) gas i...

    Text Solution

    |

  6. A sample of 16 g charcoal was brought into contact with CH(4) gas cont...

    Text Solution

    |

  7. TEL minimizes the Knocking effect when mixed with petrol. It acts as :...

    Text Solution

    |

  8. Although, nitrogen does not adsorb on a surface at room temperature, i...

    Text Solution

    |

  9. Which of the following gas molecules have minimum value of enthalpy of...

    Text Solution

    |

  10. In Langumir's model of adosrption of a gas on a solid surface :

    Text Solution

    |

  11. Which of the followings is not correct about colloids?

    Text Solution

    |

  12. From the given following sol how many can coagulate the haemoglobin so...

    Text Solution

    |

  13. From the given following sol how many can coagulate silicic acid sol? ...

    Text Solution

    |

  14. For the coagulation of 500mL of arsenious sulphide sol, 2mL of 1MNaCl ...

    Text Solution

    |

  15. The coagulation of 100 mL of a colloidal solution of gold is completel...

    Text Solution

    |

  16. The gold number of gelatin is 0.01 . Calculate the amount of gelatin t...

    Text Solution

    |

  17. 526.3mL of 0.5MHCl is shaken with 0.5g of activated charcoal and filte...

    Text Solution

    |

  18. In an experiment, addition of 5.0mL , of 0.006MBaCl(2) to 10.0mL of ar...

    Text Solution

    |

  19. In an adsorption experiment, a graph between log (x/m) versus log P wa...

    Text Solution

    |

  20. In a coagulation experiment, 5mL of As(2)S(3) is mixed with distilled...

    Text Solution

    |