Question

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

Answer 1

Answer:

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

Explanation:

Formula of Gold (III) chloride: $AuCl_{3}$

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

${N_{0}} =6.022 \times 10^{23}$

Using,

$n(moles)=\frac{Given\ number\ of\ particles}{N_{0}}$

$n =\frac{6.02\times 10^{23}}{6.022\times 10^{23}}$

= 1 mole(0.9999 , nearly equal to 1 )

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

$6.022 \times 10^{23}$  = 1 mole of $AuCl_{3}$

In this Sample,

1 mole of $AuCl_{3}$ will give = 3 mole of Chlorine atoms

1 mole of Cl contain = $6.022 \times 10^{23}$

3 mole of Cl contain = $6.022 \times 10^{23}\times 3$

3 mole of Cl contain =$18.066 \times 10^{23}$

So,

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