Question

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

Answer 1

The number of chlorine atoms present in
`6.02 * 10^(23)` units of gold III chloride is
`18.066 * 10^(23)`

Step-by-step explanation:

Formula of Gold (III) chloride:
`AuCl_(3)`

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

`{N_(0)} =6.022 * 10^(23)`

Using,

`n(moles)=(Given\ number\ of\ particles)/(N_(0))`

`n =(6.02* 10^(23))/(6.022* 10^(23))`

= 1 mole(0.9999 , nearly equal to 1 )

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

`6.022 * 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 * 10^(23)`

3 mole of Cl contain =
`6.022 * 10^(23)* 3`

3 mole of Cl contain =
`18.066 * 10^(23)`

So,

The number of chlorine atoms present in
`6.02 * 10^(23)` units of gold III chloride is
`18.066 * 10^(23)`