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If 4 g O(2) effuse through a very narrow...

If 4 g `O_(2)` effuse through a very narrow hole , How much `H_(2)` would have effused under identical conditions .

A

16 g

B

1 g

C

0.25 g

D

64 g

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The rate of effusion of an equilibrium mixture in 1 litre vessel of, at 300 K , A_(2)hArr2A through a pinhole is 0.707 times of rate of diffusion on O_(2) under identical conditions of P and T . Which of the following is/are correct if at. Wt. of A is 46 ?

Which gas effuses fastest under identical conditions?

Knowledge Check

  • If 4g of oxygen diffuse through a very narrow hole, how much hydrogen would have diffused under identical conditions ?

    A
    16g
    B
    1g
    C
    1/4 g
    D
    64 g

  • If 4g of oxygen diffuse through a very narrow hole, how much hydrogen would have diffused under identical conditions ?

    A
    16g
    B
    1g
    C
    1/4 g
    D
    64 g

  • If 4g of oxygen diffuses through a very narrow hole, how much hydrogen would have diffused under identical conditions?

    A
    16 g
    B
    1 g
    C
    `1//4` g
    D
    64 g

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    50 mL of gas A effuses through a pin hole in 146 seconds. The same volume of CO_(2) under identical condition effuses in 115 seconds. The molar mass of A is

    50 mL of gas A effuses through a pin hole in 146 seconds. The same volume of CO_2 under identical condition effuses in 115 seconds. The molar mass of A is

    The process by which a gas passes through a small hole into vacuum is called effusion. The rate of change of pressure(p) of a gas at constant temperature due to effusion of gas from a vessel of constant volume can be related to rate of change of number of molecules by the expression: (dp)/(dt)=(kT)/V((dN)/(dt)) where rate of change of number of molecules Rightarrow -((dN)/(dt))=(pA_(0))/(2pimkT)^(1//2) where k=Boltzmann constant N_(A) = Avogadro's number T= Temperature (in K) V= volume of vessel N=Number of molecules A_(0) = Area of aperture m=Mass of single molecule gamma=V/A_(0)sqrt((2pim)/(kT) If 2 g of SO_(2) effuses from given container in 10 sec then, mass of He effusing out in 30 seconds under identical conditions will be:

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    If 0.024 L of H_(2) gas is formed at the cathode, the volume of O_(2) gas formed at the anode under identical conditions, is:

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