An ancient coin has a mass of 8.630 g in air and an apparent mass of 7.800 g when fully submerged in water of density 1.000 g/cm^3. What is the density of the coin?

1. A submerged object suffers an apparent mass loss equal to the mass of the fluid it displaces. 2. Apparent mass loss = 8.630 g minus 7.800 g = 0.830 g of water displaced. 3. Volume of that water = mass / density = (0.830 g) / (1.000 g/cm^3) = 0.830 cm^3. 4. The coin is fully submerged, so 0.830 cm^3 is also the volume of the coin. 5. Density of the coin = (8.630 g) / (0.830 cm^3) = 10.4 g/cm^3, which is close to pure silver. 6. The value 0.830 g/cm^3 reports the displaced mass as though it were a density. 7. The value 1.11 g/cm^3 divides the two masses, and 19.3 g/cm^3 is the density of gold, assumed rather than measured. _Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.7 Archimedes' Principle_