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24 March, 06:17

3) At STP, a 22.4-liter sample of NH3 (g) contains the same number of molecules as A) 11.2 L of H2 (g) B) 22.4 L of CO2 (g) C) 33.6 L of CH4 (g) D) 44.8 L of O2 (g)

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  1. 24 March, 07:25
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    B) 22.4L of CO2 (g)

    Explanation:

    Given data

    At STP,

    the volume of each gas is 22.414 dm³ (1 dm³ = 1 L),

    one mole of a gas = 22.414 dm³ = 22.414 litre,

    according to Avagadro's number:

    one mole of a gas = 6.02 x 10^23 molecules.

    Solution:

    One litre of NH3 = (6.02x 10^23) / (22.414)

    =2.68 x10^22 molecules of NH3.

    22.4 litre of NH3 = 22.4 x 2.68 x10^22

    = 6.01x 10^23 molecules of NH3.

    A)

    One litre of H2 = (6.02x 10^23) / (22.414)

    =2.68 x10^22 molecules of H2.

    11.2 litre of H2 = 11.2x 2.68 x10^22

    = 3.01x 10^23 molecules of H2.

    Similarly,

    B)

    One litre of CO2 = (6.02x 10^23) / (22.414)

    =2.68 x10^22 molecules of CO2.

    22.4 litre of CO2 = 22.4 x 2.68 x10^22

    = 6.01x 10^23 molecules of CO2.

    From here we see that option B is the correct one.
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