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]
