Determine the volume of a gold bar containing the same number of atoms as an iron bar with a volume

Determine the volume of a gold bar containing the same number of atoms as an iron bar with a volume of 1 dm3 cubic. The density of gold p1 is 19.3 * 10 ^ 3 kg / m 3, the density of iron is p2-7.8 * 10 ^ 3 kg / m3

Let us find the mass and amount of iron in an ingot with a volume of 1 dm3:

m (Fe) = V (Fe) * p (Fe) = 0.001 * 7.8 * 103 = 7.8 kg;

n (Fe) = m (Fe) / M (Fe) = 7.8 / 56 = 0.139 kmol;

Let’s find the mass and volume of the gold bar:

n (Au) = n (Fe) = 0.139 kmol;

m (Au) = n (Au) * M (Au) = 0.139 * 197 = 27.439 kg;

V (Au) = m (Au) / p (Au) = 27.439 / (19.3 * 103) = 1.42 * 10-3 m3 = 1.42 dm3.

Answer: V (Au) = 1.42 dm3.