How many moles are in 1.35 x10^23 representative particles of Fluorine?

Atomic weight of fluorine is 19 g/mol.

We know that, 1 mol (19 g) of Fluorine will have Avagadro number (6.022 X 10^23) of Fluorine atoms

Thus, we have 6.022 X 10^23 atoms of F ≡ 19 g of F

Therefore, 1.35 x10^23 atoms of F ≡ 4.259 g of F

Now, 19 g of Fluorine ≡ 1 mole

∴, 4.259 g of Fluorine ≡ 0.224 mole

Answer: 0.224 moles are in 1.35 x10^23 representative particles of Fluorine

Answer: 0.224 moles

Explanation:

1) Avogadro's number is 1 mol = 6.022 x 10²³, which means that there are 6.022 x 10²³ particles in 1 mol.

2) Divide the number of of particles by Avogadro's number to find how many moles that is:

(1.35x10²³ / (6.022x10²³) = 0.224 moles