Gaseous neon has a density of 0.900 g/L at standard conditions. How many neon atoms are in 1.00 L of neon gas at standard conditions?

Answer: 1.348 * 10^23 atoms

Explanation:

Given that volume = 1.00L

At standard condition, the volume of a gas is 22.4L/mol (at S. T. P)

Volume = mole / volume at STP

1 = mole/22.4

Mole = 22.4mol.

Also

Mole = number of atoms / Avogradro constant

Where avogrado's constant = 6.02*10²³

22.4 = number of atoms/6.02*10²³

Number of atoms = 1.348*10^25atoms