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14 January, 00:38

How many grams are in 6.78 x 10^24 atoms of He?

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Answers (2)
  1. 14 January, 03:47
    0
    45.05g

    Explanation:

    Based on the Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 atoms.

    From the above findings,

    1 mole of He also contains 6.02x10^23 atoms.

    1 mole of He = 4g

    If 4g of He contains 6.02x10^23 atoms,

    then Xg of He will contain 6.78x10^24 atoms i. e

    Xg of He = (4x6.78x10^24) / 6.02x10^23

    Xg of He = 45.05g

    Therefore, 45.05g of He contains 6.78x10^24 atoms
  2. 14 January, 03:52
    0
    45.05grams

    Explanation:

    Number of atoms present in a substance = number of moles of the substance * Avogadro's number

    Given, number of atoms = 6.78*10^24

    Number of moles = mass / molar mass

    Molar mass of Helium = 4g/mol

    Avogadro's number = 6.02*10^23

    Therefore,

    6.78*10^24 = (m/4) * (6.02*10^23)

    6.78*10^24 = (6.02*10^23*m) / 4

    Cross multiply

    6.02*10^23*m=4*6.78*10^24

    Divide both sides by 6.02*10^23

    m = (4*6.78*10^24) / 6.02*10^23

    m = (27.12*10^24) / 6.02*10^23

    m=45.05grams

    Therefore, there are 45.05grams in 6.78*10^24 atoms of helium
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