Home
Class 11
CHEMISTRY
The ratio of number of oxygen atoms (O) ...

The ratio of number of oxygen atoms `(O)` in `16.0g` ozone `(O_(3)) . 28.0g` carbon monoxide `(CO)` and `16.0g` oxygen `(O_(2))` is :-
`("Atomic mass" : C=12,O=16 "and Avogadro's constant" N_(A) =6.0xx10^(23) "mol"^(-1))`

A

`3:1:1`

B

`1:1:2`

C

`3:1:2`

D

`1:1:1`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to calculate the number of oxygen atoms in each of the three substances: ozone (O₃), carbon monoxide (CO), and oxygen (O₂). We will then find the ratio of these quantities. ### Step 1: Calculate the number of moles of ozone (O₃) - **Given:** Mass of ozone = 16.0 g - **Molar mass of ozone (O₃)** = 3 × atomic mass of oxygen = 3 × 16 g/mol = 48 g/mol - **Number of moles of O₃ (n)** = mass / molar mass = 16.0 g / 48 g/mol = 1/3 mol ### Step 2: Calculate the number of oxygen atoms in ozone (O₃) - Each molecule of O₃ contains 3 oxygen atoms. - **Number of oxygen atoms in O₃** = moles of O₃ × number of oxygen atoms per molecule × Avogadro's number = (1/3 mol) × 3 × (6.022 × 10²³ atoms/mol) = 6.022 × 10²³ atoms ### Step 3: Calculate the number of moles of carbon monoxide (CO) - **Given:** Mass of CO = 28.0 g - **Molar mass of CO** = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol - **Number of moles of CO (n)** = mass / molar mass = 28.0 g / 28 g/mol = 1 mol ### Step 4: Calculate the number of oxygen atoms in carbon monoxide (CO) - Each molecule of CO contains 1 oxygen atom. - **Number of oxygen atoms in CO** = moles of CO × number of oxygen atoms per molecule × Avogadro's number = (1 mol) × 1 × (6.022 × 10²³ atoms/mol) = 6.022 × 10²³ atoms ### Step 5: Calculate the number of moles of oxygen (O₂) - **Given:** Mass of O₂ = 16.0 g - **Molar mass of O₂** = 2 × atomic mass of oxygen = 2 × 16 g/mol = 32 g/mol - **Number of moles of O₂ (n)** = mass / molar mass = 16.0 g / 32 g/mol = 1/2 mol ### Step 6: Calculate the number of oxygen atoms in oxygen (O₂) - Each molecule of O₂ contains 2 oxygen atoms. - **Number of oxygen atoms in O₂** = moles of O₂ × number of oxygen atoms per molecule × Avogadro's number = (1/2 mol) × 2 × (6.022 × 10²³ atoms/mol) = 6.022 × 10²³ atoms ### Step 7: Find the ratio of the number of oxygen atoms Now we have: - Number of oxygen atoms in O₃ = 6.022 × 10²³ atoms - Number of oxygen atoms in CO = 6.022 × 10²³ atoms - Number of oxygen atoms in O₂ = 6.022 × 10²³ atoms Thus, the ratio of the number of oxygen atoms in O₃ : CO : O₂ is: 1 : 1 : 1 ### Final Answer The ratio of the number of oxygen atoms in 16.0 g ozone (O₃), 28.0 g carbon monoxide (CO), and 16.0 g oxygen (O₂) is **1:1:1**. ---
Promotional Banner

Topper's Solved these Questions

  • MOLE CONCEPT

    ALLEN|Exercise Exercise - 05(B)|8 Videos
  • MOLE CONCEPT

    ALLEN|Exercise Exercise - 04[B]|27 Videos
  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|7 Videos
  • QUANTUM NUMBER & PERIODIC TABLE

    ALLEN|Exercise J-ADVANCED EXERCISE|24 Videos

Similar Questions

Explore conceptually related problems

Calculate the number of oxygen atoms in 0.20 mole of Na_(2)CO_(3).10H_(2)O .

Find the ratio of the number of atoms present in 16 g of O_2 and 32 g of O_3.

The number of oxygen atoms present in 20.4 g of Al_(2)O_(3) are equal to the number of :

The number of oxygen atoms in 4.4 of CO_(2) is (Given that atomic mass C and O are 12 and 16 g/mol)

Calcuate the mass of 10^(22) atoms of sulphur. [ Atocmi mass S=32,C and O=16 and Avogadro's number =6xx10^(23) ]

Total number of moles of oxygen atoms in 3 litre O_3 (g) at 27^@C and 8.21 atm are :

The mass of an atom of oxygen is 16.0 u. Express it in kg.

The total number of electrons in 1.6 g of CH_(4) to that in 1.8 g of H_(2)O

Calculate the mass of 0.1 mole of carbon dioxide. [Atomic mass : S = 32, C = 12 and O = 16 and Avogadro.s Number = 6 xx 10^(23) ]

Find number of oxygen atoms present in 100 mg of CaCO_3. (Atomic mass of Ca = 40 u, C = 12 u, O = 16 u)

ALLEN-MOLE CONCEPT-Exercise - 05(A)
  1. In a compound C, H, N atoms are present in 9:1:3.5 by weight. Molecula...

    Text Solution

    |

  2. 6.02xx10^(23) molecules of urea are present in 100 ml of its solution....

    Text Solution

    |

  3. If 1//6, in place of 1//12, mass of carbon atom is taken to be the rel...

    Text Solution

    |

  4. How many moles of magnesium phosphate Mg3 (PO4)2 will contain 0.25 mol...

    Text Solution

    |

  5. In the reaction : 2Al(s) + 6HCl(aq) rarr 2 Al^(3+)(aq) + 6 Cl^(-)(aq) ...

    Text Solution

    |

  6. If 10^(-4) dm^(3) of water is introduced into a 1.0dm^(3) flask at 300...

    Text Solution

    |

  7. A 5.2 molal aqueous solution of methyl alcohol, CH(3)OH is supplied....

    Text Solution

    |

  8. The mass of potassium dichromate crystals required to oxidize 750 cm^(...

    Text Solution

    |

  9. A transition metal M forms a volatile chloride which has a vapour dens...

    Text Solution

    |

  10. The ratio of number of oxygen atoms (O) in 16.0g ozone (O(3)) . 28.0g ...

    Text Solution

    |

  11. The density of a solution prepared by dissolving 120 g of urea ( mol. ...

    Text Solution

    |

  12. The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl wi...

    Text Solution

    |

  13. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g...

    Text Solution

    |

  14. For the estimation of nitrogen 1.4 g of organic compound was diagest b...

    Text Solution

    |

  15. The ratio of masses of oxygen and nitrogen in a particular gaseous mix...

    Text Solution

    |

  16. The molecular formula of a commercial resin used for exchanging ions i...

    Text Solution

    |

  17. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air...

    Text Solution

    |

  18. The amount of arsenic pentasulphide that can be obtained when 35.5 g a...

    Text Solution

    |

  19. An organic compound contains C, H and S. The minimum molecular weight ...

    Text Solution

    |

  20. 5L of an alkane requires 25L of oxygen for its complete combustion. If...

    Text Solution

    |