Home
Class 11
CHEMISTRY
X mL of H(2) gas effuses through a hole ...

`X mL` of `H_(2)` gas effuses through a hole in a container in `5 s`. The time taken for the effusion of the same volume of the gas specified below, under identical conditions, is

A

10 sec : He

B

20 sec : `O_(2)`

C

25 sec : CO2

D

55 sec : `CO_(2)`

Text Solution

Verified by Experts

The correct Answer is:
b
(b) `(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))`
`(V_(1))/(t_(1))xx(t_(2))/(V_(2))=sqrt((M_(2))/(M_(1)))`
Promotional Banner

Topper's Solved these Questions

  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 1 (Q.151 To Q.176)|1 Videos

  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 2|1 Videos

  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 1 (Q.121 To Q.150)|1 Videos

  • ELECTROCHEMISTRY

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Assertin-Reason Type Questions|1 Videos

Similar Questions

Explore conceptually related problems

The molecular weight of a gas which diffuses through a porous plug at 1//6^(th) of the speed of hydrogen under identical condition is:

Under identical conditions of pressure and temperature, 4 L of gaseous moxture ( H_(2) and CH_(4) ) effuses through a hole in 5 min whereas 4 L of a gas X of molecular mass 36 takes to 10 min to effuse through the same hole. The mole ratio of H_(2):CH_(4) in the mixture is:

A given volume of ozonised oxygen (containing 60% oxygen by volume ) required 220 sec to effuse while an equal volume of oxygen took 200 sec only under identical conditions. If density of O_(2) is 1.6 g/L then find density of O_(3) .

A mixture of O_(2) and gas "Y" ( mol. wt. 80) in the mole ratio a:b has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio b:a under identical conditions ? ( gases are )

The time for a certain volume of a gas A to diffuse through a small hole is 2 minute If takes 5.65 minute for oxygen to diffuse under similar conditions Find the molecualr weight of A .

At 800^(@)C, the following equilibrium is established as F_(2)(g)hArr2F(g) The cojmpositionof equilibrium may be determinded by measuring the rate of effusion of theh kmixture through a pin hole. It is found that at 800^(@)C and 1 atm mixture effuses 1.6 times as fast as SO_(2) effuse under the similar conditions. (At. mass of F = 19 ) what is the value of K_(p) (in atm) ?

A rigid container contains 5 mole H_(2) gas at some pressure and temperature. The gas has been allowed to escape by simple process from the container due to which pressure of the gas becomes half of its initial pressure and temperature become (2//3)^(rd) of its initial. The mass of gas remaining is :

MnO_4^(-) ions are reduced in acidic condition to Mn^(2+) ions whereas they are reduced in neutral condition to MnO_2 . The oxidation of 25 ml of a solution X containing Fe^(2+) ions required in acidic condition 20 ml of a solution Y containing MnO_4^(-) ions. What volume of solution Y would be required to oxidise 25 ml of solution X containing Fe^(2+) ions in neutral condition ?

The existence of negatively charged particle in an atom was shown by J.J. Thomson as a result of the studies of the passage of electricity through gases at extremely low pressure known as discharge tube experiments. When a high voltage of the order of 10,000 volts or more was impressed across the electrodes, some sort of invisible rays moved from the negative electrode to the positive electrode these rays are called as cathode rays. Cathode rays travel in straight path in absence of electrical and magnetic field . Cathode rays consist of material part and charged particles? Cathode rays produce X-rays and light is emitted when they strike on ZnS screen. Cathode rays penetrate through thin sheets of aluminium and other metals . They affect the photogenic plate and passes heating effect when they strike on metal foil. The raito of charge to mass i.e charge/mass is same for all the cathode rays irrespective of the gas used in the tube. The existence of positively charged particle in an atom was shown be E. Goldstein. He repeated the same discharge tube experiments by using a perforated cathode. It was observed that when a high potential difference was applied between the electrodes, not only cathode rays were produced but also a new type of rays were produced simultaneoulsy from anode moving towards cathode and passes through the holes or canal of the cathode. These termed as canal rays or anode rays. These rays travel in straight lines and consists of positively charged particles. These rays have kinetic energy and produces heating effect also. The e/m ratio of these rays is smaller than that of electrons. Unlike cathode rays, their e/m value is dependent upon the nature of the gas taken in the tube. These rays produced flashes of light on ZnS screen and can pass throughs thin metal foils. They can produce physical and chemical changes and are capable to produce ionisation in gases. For cathode rays the value of e/m:

NARENDRA AWASTHI-GASEOUS STATE-Exercise
  1. What is the ratio of diffusion rate of oxygen to hydrogen?

    Text Solution

    |

  2. The molecular weight of a gas which diffuses through a porous plug at...

    Text Solution

    |

  3. X mL of H(2) gas effuses through a hole in a container in 5 s. The tim...

    Text Solution

    |

  4. At identical temperature and pressure the rate of diffusion of hydroge...

    Text Solution

    |

  5. Calculate relative rate of effusion of O(2) to CH(4) from a container ...

    Text Solution

    |

  6. The following organic compound are popularly known by their common nam...

    Text Solution

    |

  7. The vapour density of a mixture contaninig NO(2) and N(2)O(4) is 27.6...

    Text Solution

    |

  8. 80 mL of O(2) takes 2 minutes to pass through the hole. What volume of...

    Text Solution

    |

  9. When two cotton plugs, one moistened with ammonia and the other with h...

    Text Solution

    |

  10. 4 gm of sulphur dioxide gas diffuses from a container in 8 min. Mass o...

    Text Solution

    |

  11. Under identical conditions of pressure and temperature, 4 L of gaseou...

    Text Solution

    |

  12. A ballon weighing 50 kg is filled with 685kg of helium at 1atm pressur...

    Text Solution

    |

  13. According to the law of equipartition of energy, the energy associated...

    Text Solution

    |

  14. Calculate gamma (ratio of C(p) and C(v)) for triatomic linear gas at h...

    Text Solution

    |

  15. If one mole of a mono-atomic gas (gamma=5//3) is mixed with one mole o...

    Text Solution

    |

  16. If one mole of a monatomic gas (gamma=5/3) is mixed with one mole of a...

    Text Solution

    |

  17. Select the correct statement :

    Text Solution

    |

  18. Which gas shows real behaviour?

    Text Solution

    |

  19. Consider an ideal gas contained in a vessel If the intermolecular inte...

    Text Solution

    |

  20. The pressure of real gas is less than the pressure of an ideal gas bec...

    Text Solution

    |