Gold can be flattened to a thickness of 0.1 microns. What surface area can be covered with a face of gold

Gold can be flattened to a thickness of 0.1 microns. What surface area can be covered with a face of gold, the mass of which is m = 2.0 g?

Given:
m = 2 grams = 0.002 kilograms – the mass of gold;
ro = 19320 kg / m ^ 3 – density of pure gold;
h = 0.1 micrometer = 1 * 10 ^ -7 meters – the thickness to which gold can be flattened.
It is required to determine S (m ^ 2) – the surface area that can be covered with gold of mass m.
Find the volume of gold:
V = m / ro = 0.002 / 19320 = 1 * 10 ^ -7 m ^ 3.
Then the surface area will be equal to:
S = V / h = 1 * 10 ^ -7 / 10 ^ -7 = 1 m ^ 2.
Answer: gold weighing 0.002 kilograms can cover a surface with an area of 1 square meter.