Density: Question 5

Worked solution

Part (a): Mass of each metal

Rearrange the density formula, $\rho = \dfrac{m}{V}$, to make mass the subject: $m = \rho V$.

Metal A:

$m_A = 8.90\text{ g/cm}^3 \times 50.0\text{ cm}^3 = 445\text{ g}$

Metal B:

$m_B = 7.14\text{ g/cm}^3 \times 30.0\text{ cm}^3 = 214.2\text{ g}$

Part (b): Density of the alloy

The alloy’s density needs the total mass divided by the total volume, not an average of the two original densities.

Total mass:

$m_{\text{total}} = m_A + m_B = 445 + 214.2 = 659.2\text{ g}$

Total volume:

$V_{\text{total}} = 50.0 + 30.0 = 80.0\text{ cm}^3$

Density of the alloy:

$\rho_{\text{alloy}} = \frac{m_{\text{total}}}{V_{\text{total}}} = \frac{659.2\text{ g}}{80.0\text{ cm}^3}$

$\rho_{\text{alloy}} = 8.24\text{ g/cm}^3$

As a check, this value lies between the two component densities ($7.14$ and $8.90\text{ g/cm}^3$), closer to metal A since a greater volume of A was used, as expected for a mass-weighted combination.

Part (c): Floating or sinking in water

Compare the alloy’s density with the density of water:

$8.24\text{ g/cm}^3 \gg 1.00\text{ g/cm}^3$

Since the alloy is much denser than water, it sinks.

Final answers

• (a) $m_A = \mathbf{445\text{ g}}$, $m_B = \mathbf{214.2\text{ g}}$
• (b) Density of the alloy $=$ $8.24\text{ g/cm}^3$
• (c) The alloy sinks in water.