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22 August, 18:34

What is the atomic weight of an element if 4 grams of it contains 2.98 x 10^22 atoms?

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Answers (2)
  1. 22 August, 18:44
    0
    N=amount in moles

    N=amount in numbers=2.98x10^22

    L=Avagadro's constant=mole-1 (6.022x10^23)

    m = mass in grams

    M=molar mass in g/mol

    Now,

    n=N/L

    =2.98x10^22/6.022x10^23

    =0.0495 mol ≈0.05 mol

    Then,

    M=m/n

    =4/0.05

    =80 g/mol

    The element is Br
  2. 22 August, 20:25
    0
    M = 4 g; N = 2.98 * 10^22

    Avogadro constant: L = 6.022 * 10^23

    Number of moles:

    n = 2.98 * 10^22 : 6.022 * 10^23 = 0.298 * 10^23 : 6.022 * 10^23 =

    = 0.0495 moles

    Molar mass:

    M = m/n = 4 g / 0.0495 = 80.8 ≈ 80
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