Question

A single gold atom has a mass of 197.0 amu. How many gold atoms are in a cube of gold that is 10.0 mm on each side (about the size of a sugar cube) if the density of gold is 19.3 g/cm3

Answer 1

Answer:

5.90x10²² atoms of gold are present in the cube

Explanation:

First, we must find the volume of the cube in cm³. With density we can find the mass of the gold and the moles using its molar mass. As 1 mol = 6.022x10²³ atoms we can find the number of atoms:

Volume in cm³:

(10.0mm)³ = 1000mm³

1mm³ = 0.001cm³

1000mm³ * (0.001cm³ / 1mm³) = 1cm³

Mass gold:

1cm³ * (19.3g/cm³) = 19.3g Gold

Moles Gold:

19.3g * (1mol / 197.0g) = 0.0980 moles Gold

Atoms gold:

0.0980 moles Gold * (6.022x10²³ atoms / mol) =

5.90x10²² atoms of gold are present in the cube