Question

How many neon atoms in a `15.8*g` mass of neon gas?

Answer 1

See below..

Explanation:

No. of moles `=("Given mass")/("Atomic Mass")`= `15.8/20=0.79`

No. of atoms = (Avogadro's no.) `*`(no. of moles)=
`(6.022 *10^23 *0.79)=4.75*10^23`

Answer 2

Well, we know that there are `6.022xx10^23` particles in a mole of such particles....

Explanation:

And we also know (do we?) that a `20.18*g` mass of neon gas contains `6.022xx10^23` individual neon atoms....

And so we work out the quotient multiplied by the `"Avocado number...."`

`(15.8*g)/(20.18*g*mol^-1)xx6.022xx10^23*mol^-1`

`~=4.5xx10^23*"individual neon atoms"`...

Here I have used `"Avogadro's number"` precisely as I would any other collective number: `"dozen"`; `"Bakers' dozen"`; `"score"`; `"gross"`...

We use `"Avogadro's number"` because we know that such a number of neon atoms has a mass of `20.18*g`...

Answer 3

`4.71 * 10^23` atoms

Explanation:

Neon has an atomic mass of 20.18. Divide 15.8 grams by the atomic mass to determine the moles of Ne. Then multiply the moles of Ne by Avogadro's number to get the number of atoms.

`(15.8g -: 20.18` moles`) 6.02 * 10^23` atoms = `4.71 * 10^23` atoms