Density

Density is mass per unit volume.

Finding the volume of the cube $$V=(2.0{)}^{3}=8.0{\text{ cm}}^{3}$$ Converting volume to SI units $$V=\frac{8.0}{{10}^{6}}$$ $$V=8.0\times {10}^{-6}{\text{ m}}^{3}$$ Converting mass to SI units $$m=\frac{24}{1000}$$ $$m=0.024\text{ kg}$$ Equation used — density equation $$\rho =\frac{m}{V}$$ Substituting $$\rho =\frac{0.024}{8.0\times {10}^{-6}}$$ $$\rho =3000{\text{ kg m}}^{-3}$$

Measure the mass of the empty measuring cylinder on the balance; pour a known volume of cooking oil into the cylinder and read the volume at the bottom of the meniscus at eye level; measure the new mass of cylinder plus oil; subtract to find the mass of oil alone; calculate density using ρ = m/V.

Finding the mass of the liquid $$m=157-85$$ $$m=72\text{ g}$$ Equation used — density equation $$\rho =\frac{m}{V}$$ Substituting $$\rho =\frac{72}{60}$$ $$\rho =1.2{\text{ g cm}}^{-3}$$

Finding the volume of the cylinder $$V=\pi {r}^{2}h$$ Substituting $$V=\pi \times (1.5{)}^{2}\times 8.0$$ $$V=56.5{\text{ cm}}^{3}$$ Equation used — density equation $$\rho =\frac{m}{V}$$ Substituting $$\rho =\frac{425}{56.5}$$ $$\rho =7.52\approx 7.5{\text{ g cm}}^{-3}\text{ (2 s.f.)}$$

A ruler gives a direct measurement of each dimension, so the volume can be calculated precisely from the mathematical formula; the displacement method relies on reading water levels, which introduces parallax error and uncertainty from water drops clinging to the solid; therefore the ruler method gives a more accurate volume for a regular shape.

Pour water into a measuring cylinder and record the water level ($V_{1}$) at the bottom of the meniscus at eye level; gently lower the stone into the water so it is fully submerged; record the new water level ($V_{2}$); the volume of the stone equals ($V_{2}$ - $V_{1}$).

Finding the volume of the stone $$V=100.0-40.0$$ $$V=60.0{\text{ cm}}^{3}$$ Equation used — density equation $$\rho =\frac{m}{V}$$ Substituting $$\rho =\frac{156}{60.0}$$ $$\rho =2.60{\text{ g cm}}^{-3}$$ Converting to kg m⁻³ $$\rho =2.60\times 1000$$ $$\rho =2600{\text{ kg m}}^{-3}$$

The iron ball floats; because the density of iron (7 800 kg m⁻³) is less than the density of mercury (13 600 kg m⁻³), so the upward buoyant force from the mercury can support the weight of the iron ball.

The liquid with the lowest density (790 kg m⁻³) floats on top; the liquid with the middle density (920 kg m⁻³) forms the middle layer; the liquid with the highest density (1050 kg m⁻³) sinks to the bottom. Therefore the order from top to bottom is: 790 kg m⁻³, 920 kg m⁻³, 1050 kg m⁻³. This occurs because, for immiscible liquids, the less dense liquid always floats above the denser liquid.

Reading at the bottom of the meniscus gives the true volume of water, because the surface curves upward at the edges due to adhesion to the glass; reading at eye level prevents parallax error, which occurs when viewing at an angle and causes the reading to appear higher or lower than the true value.

Finding the volume of the stone $$V=73.5-50.0$$ $$V=23.5{\text{ cm}}^{3}$$ Equation used — density equation $$\rho =\frac{m}{V}$$ Substituting $$\rho =\frac{62.1}{23.5}$$ $$\rho =2.64{\text{ g cm}}^{-3}\text{ (3 s.f.)}$$ The student’s result (2.64 g cm⁻³) is close to but slightly below the known value (2.65 g cm⁻³). One reason is that air bubbles may have been trapped on the surface of the stone when it was submerged, increasing the apparent volume of displaced water and therefore giving a slightly lower calculated density.