DENSITY

1. Mass and Volume

  • Mass is the amount of matter in a material; all matter has mass.

  • Mass is measured in kilograms (kg) or grams (g).

  • The SI unit for mass is the kilogram (kg).

  • Apparatus commonly used to measure mass: triple beam balance, mass meter, electronic scale.

2. Volume

  • Volume can be measured with: measuring cylinder, graduated beaker, measuring flask, burette.

  • For solids, volume units include: m³, cm³, mm³.

  • For liquids, volume units include: mL and L.

  • Volume formula for a rectangular solid: V=l×w×hV = l \times w \times h.

  • The volume of an irregular object can be found by water displacement (see Section 1.4).

  • 1 mL = 1 cm³ = (1 cm × 1 cm × 1 cm).

  • 1 L = 1000 cm³ = (10 cm × 10 cm × 10 cm).

3. Density

  • Density of a material is the amount of mass in a given volume of that material.

  • Formula: density=massvolume=mV\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{m}{V}

  • Common density units include: \text{g mL}^{-1} \text{ (or g·mL}^{-1}), g cm3\text{g cm}^{-3}, kg L1\text{kg L}^{-1}, etc.

  • The unit for density depends on the mass and volume units used.

  • Density units table (variables):

    • mm: mass

    • VV: volume

    • dd: density

    • Density symbol: dd

    • Units examples: g mL1,  g cm3,  kg L1\text{g mL}^{-1},\; \text{g cm}^{-3},\; \text{kg L}^{-1}

  • Important note on units: for density calculations, mass and volume units must be compatible. If mass is in g, volume should be in mL or cm³; if mass is in kg, volume should be in L.

  • Key relationships:

    • d=mVd = \frac{m}{V}

    • V=mdV = \frac{m}{d}

    • m=d×Vm = d \times V

4. Factors that affect density

  • The density of a material depends on:

    • The nature of the particles and the strength of the forces between particles.

    • The size and type of the particles.

    • The size of the spaces (voids) between particles.

  • Therefore, different materials with different internal structures will have different densities.

5. Density Calculations (worked examples)

  • Example 1: If a 10 g object occupies 2.5 cm³, what is its density?

    • d=mV=10g2.5cm3=4g cm3d = \frac{m}{V} = \frac{10\,\text{g}}{2.5\,\text{cm}^3} = 4\,\text{g cm}^{-3}

  • Example 2: What is the mass of methanol that fills exactly a 200 mL container if the density is 0.789 g mL10.789\ \text{g mL}^{-1}?

    • m=d×V=0.789×200=157.8 gm = d \times V = 0.789 \times 200 = 157.8\ \text{g}

  • Example 3: Copper block density

    • Mass m=1896 gm = 1896\ \text{g}.

    • Dimensions: l=8.4 cm,  w=5.5 cm,  h=4.6 cml = 8.4\ \text{cm},\; w = 5.5\ \text{cm},\; h = 4.6\ \text{cm}

    • Volume: V=l×w×h=8.4×5.5×4.6=212.52 cm3V = l \times w \times h = 8.4 \times 5.5 \times 4.6 = 212.52\ \text{cm}^3

    • Density: d=mV=1896212.528.92 g cm3d = \frac{m}{V} = \frac{1896}{212.52} \approx 8.92\ \text{g cm}^{-3}

6. Measuring density of irregular objects (water displacement method)

  • Principle: When an object is submerged in water, it displaces an amount of water equal to its volume.

  • This method is especially useful for irregularly shaped objects (e.g., rocks).

  • Procedure (illustrative):

    • Fill a graduated cylinder with water (record initial volume).

    • Submerge the irregular object fully.

    • Record the new water level (final volume).

    • Volume of object = final volume − initial volume.

  • Example 4: Rock volume by displacement

    • Initial water: 50 mL; final water after submersion: 62 mL.

    • Volume of rock = 62 − 50 = 12 mL = 12 cm³.

  • Example 5: Rock density with mass and volume

    • Volume = 47 mL; Mass = 78 g.

    • Density: d=mV=78471.66 g mL1d = \frac{m}{V} = \frac{78}{47} \approx 1.66\ \text{g mL}^{-1}

7. Practice problems and worksheets (summary of problem types and solutions)

  • Worksheet 1 (types of problems): density calculations using d = m/V and V = m/d with various data; checks on understanding that larger density implies more mass for the same volume.

  • Example 1 (from Worksheet 1):

    • Given mass 500 g and final volume 555 mL, density d=mV=5005550.90 g mL1d = \frac{m}{V} = \frac{500}{555} \approx 0.90\ \text{g mL}^{-1}

  • Example 2 (Chalk):

    • Mass 50 g, density 2.6 g/mL, volume V=md=502.619.23 mLV = \frac{m}{d} = \frac{50}{2.6} \approx 19.23\ \text{mL}

  • Example 3 (Coal vs Cork):

    • Given a volume of 100 mL, densities: coal d<em>coal=0.25 g mL1d<em>{coal} = 0.25\ \text{g mL}^{-1} and cork d</em>cork=1.50 g mL1d</em>{cork} = 1.50\ \text{g mL}^{-1}

    • Masses: m<em>coal=d</em>coal×V=0.25×100=25 gm<em>{coal} = d</em>{coal} \times V = 0.25 \times 100 = 25\ \text{g}

    • m<em>cork=d</em>cork×V=1.50×100=150 gm<em>{cork} = d</em>{cork} \times V = 1.50 \times 100 = 150\ \text{g}

    • Conclusion: For identical volumes, the denser material has a larger mass (coal vs cork in this example).

  • Worksheet 2 (additional problems): varied values for density calculations, including

    • Example: d=mV,  m=12.9 g,  V=8 cm3d=12.98=1.61 g cm3d = \frac{m}{V},\; m = 12.9\ \text{g},\; V = 8\ \text{cm}^{3} \Rightarrow d = \frac{12.9}{8} = 1.61\ \text{g cm}^{-3}

    • Example: d=mV,  m=43.5 g,  V=50 mLd=43.550=0.87 g mL1d = \frac{m}{V},\; m = 43.5\ \text{g},\; V = 50\ \text{mL} \Rightarrow d = \frac{43.5}{50} = 0.87\ \text{g mL}^{-1}

  • Worksheet 3/Examples (more practice problems):

    • Example: m=d×V,  V=15 mL,  d=2.5 g mL1m=2.5×15=37.5 gm = d \times V,\; V = 15\ \text{mL},\; d = 2.5\ \text{g mL}^{-1} \Rightarrow m = 2.5 \times 15 = 37.5\ \text{g}

    • Example: V=md,  m=65 g,  d=5.45 g mL1V=655.4511.93 mLV = \frac{m}{d},\; m = 65\ \text{g},\; d = 5.45\ \text{g mL}^{-1} \Rightarrow V = \frac{65}{5.45} \approx 11.93\ \text{mL}

    • Example: d=mV,  m=176 g,  V=4 cm3d=1764=44 g cm3d = \frac{m}{V},\; m = 176\ \text{g},\; V = 4\ \text{cm}^3 \Rightarrow d = \frac{176}{4} = 44\ \text{g cm}^{-3} (note: ensure units are consistent; this example demonstrates formula usage)

  • Worksheet 4 (more large-volume/mass figures):

    • Example: d=mV,  m=2700 g,  V=10×15×35 cm3=5250 cm3d270052500.51 g cm3d = \frac{m}{V},\; m = 2700\ \text{g},\; V = 10\times 15 \times 35\ \text{cm}^3 = 5250\ \text{cm}^3 \Rightarrow d \approx \frac{2700}{5250} \approx 0.51\ \text{g cm}^{-3}

    • Example: d=mV,  m=78 g,  V=60 mLd=7860=1.30 g mL1d = \frac{m}{V},\; m = 78\ \text{g},\; V = 60\ \text{mL} \Rightarrow d = \frac{78}{60} = 1.30\ \text{g mL}^{-1}

    • Example: d=mV,  m=1300 g,  V=743 cm3d13007431.75 g cm3d = \frac{m}{V},\; m = 1300\ \text{g},\; V = 743\ \text{cm}^3 \Rightarrow d \approx \frac{1300}{743} \approx 1.75\ \text{g cm}^{-3}

8. Practical tips and key reminders

  • Always ensure mass and volume units are compatible before calculating density.

    • If mass is in g, use volume in mL or cm³.

    • If mass is in kg, use volume in L.

  • Remember the difference between solids and liquids in how volume is reported and measured.

  • For irregular objects, water displacement gives a direct measurement of volume, which is often easier than calculating by geometry.

  • Use the formulae consistently:

    • d=mVd = \frac{m}{V}

    • V=mdV = \frac{m}{d}

    • m=d×Vm = d \times V

  • Density is a material property that helps identify substances and compare materials under the same conditions.

  • Real-world relevance: density is used in material science, quality control, buoyancy, and identification of substances. It can indicate composition and purity when measured under standard conditions.