Example 1.1: Calculation of Density
Calculation of Density
Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold,
19.3 g/cm^3. What is the density of lead if a cube of lead has an edge length of
2.00 cm and a mass of
90.7 g?
Solution
\(\mathrm{edge}_{\mathrm{cube}}\) \(= 2.00\ \mathrm{cm}\)
\(m_{\mathrm{lead}}\) \(= 90.7\ \mathrm{g}\)
The density ρ of a substance can be calculated by dividing its mass by its volume.
\(ρ_{\mathrm{substance}} = \dfrac{m_{\mathrm{substance}}}{V_{\mathrm{substance}}}\)
The volume of a cube is calculated by cubing the edge length.
\(V_{\mathrm{cube}}\) \(= {\mathrm{edge}_{\mathrm{cube}}}^{3}\)
\(\ \ \ ={(2.00\ \mathrm{cm})}^{3}\)
\(\ \ \ =8.0\ \mathrm{cm}^{3}\)
\(ρ_{\mathrm{lead}}\) \(= \dfrac{m_{\mathrm{lead}}}{V_{\mathrm{cube}}}\)
\(\ \ \ =\dfrac{90.7\ \mathrm{g}}{8.0\ \mathrm{cm}^{3}}\)
\(\ \ \ =11.3\ \frac{\mathrm{g}}{\mathrm{cm}^{3}}\)