Chemistry

Question

How many chlorine atoms would be in 6.02 X 10^23 units of gold III chloride

1 Answer

  • Answer:

    The number of chlorine atoms present in [tex]6.02 \times 10^{23}[/tex] units of gold III chloride is [tex]18.066 \times 10^{23}[/tex]

    Explanation:

    Formula of Gold (III) chloride: [tex]AuCl_{3}[/tex]

    Avogadro Number : Number of particles present in one mole of a substance.

    [tex]{N_{0}} =6.022 \times 10^{23}[/tex]

    Using,

    [tex]n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}[/tex]

    [tex]n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}[/tex]

    = 1 mole(0.9999 , nearly equal to 1 )

    The given Gold III chloride sample is 1 mole in amount.

    [tex]6.022 \times 10^{23}[/tex]  = 1 mole of [tex]AuCl_{3}[/tex]

    In this Sample,

    1 mole of [tex]AuCl_{3}[/tex] will give = 3 mole of Chlorine atoms

    1 mole of Cl contain = [tex]6.022 \times 10^{23}[/tex]

    3 mole of Cl contain = [tex]6.022 \times 10^{23}\times 3[/tex]

    3 mole of Cl contain =[tex]18.066 \times 10^{23}[/tex]

    So,

    The number of chlorine atoms present in [tex]6.02 \times 10^{23}[/tex] units of gold III chloride is [tex]18.066 \times 10^{23}[/tex]

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