CHEM 1315 0.3 - Density and Volume Practice

Key Equations and Unit Conversions

Density formula: \rho = \frac{m}{V}
Volume from density and mass: V = \frac{m}{\rho}
Mass from density and volume: m = \rho V
Volume by displacement for irregular objects: V{\text{object}} = V{\text{final}} - V_{\text{initial}}
For cubes: V = a^3 and a = \sqrt[3]{V}
Unit relationships: 1\ \text{L} = 10^3\ \text{mL},\; 1\ \text{cm}^3 = 1\ \text{mL}
Density units: g/cm$^3$ and g/mL are equivalent numerically
Significant figures guideline: answer with the same precision as the least precise input values

Given:
- Empty flask mass: m_{\text{empty}} = 78.23\ \mathrm{g}
- Flask filled with water: m_{\text{water}} = 593.63\ \mathrm{g}
- Density of water: \rho_{\text{water}} = 1.00\ \mathrm{g\,cm^{-3}}
- Flask filled with concentrated sulfuric acid: m_{\text{acid+flask}} = 1026.57\ \mathrm{g}
Steps:
- Volume of water (and thus flask capacity) from water mass:
  V{\text{flask}} = \frac{m{\text{water}} - m{\text{empty}}}{\rho{\text{water}}} = \frac{593.63 - 78.23}{1.00} = 515.40\ \mathrm{mL}
- Mass of acid in the flask:
  m{\text{acid}} = m{\text{acid+flask}} - m_{\text{empty}} = 1026.57 - 78.23 = 948.34\ \mathrm{g}
- Density of concentrated sulfuric acid:
  \rho{\text{acid}} = \frac{m{\text{acid}}}{V_{\text{flask}}} = \frac{948.34}{515.40} \approx 1.84\ \mathrm{g\,cm^{-3}}
Answer (to 3 s.f.): \rho_{\text{H2SO4}} \approx 1.84\ \mathrm{g\,cm^{-3}}
Notes:
- Uses the same flask volume for both liquids; mass difference gives liquid mass; density uses volume from water as the flask capacity

Given:
- Ring mass: m = 5.00\ \mathrm{g}
- Ring volume via water displacement: initial volume in cylinder V{i} = 10.000\ \mathrm{mL}, final volume V{f} = 10.476\ \mathrm{mL}
Steps:
- Displaced volume:
  V = V{f} - V{i} = 10.476 - 10.000 = 0.476\ \mathrm{mL}
- Density of ring:
  \rho = \frac{m}{V} = \frac{5.00}{0.476} \approx 10.5\ \mathrm{g\,cm^{-3}}
Comparison:
- Known density of silver: \rho_{\text{Ag}} \approx 10.49\ \mathrm{g\,cm^{-3}}
- Measured density ~ $10.5\ \mathrm{g\,cm^{-3}}$ is consistent with silver
Answer: The ring is consistent with being pure silver (assuming it is solid and not an alloy with similar density)
Hint discussion: What information would you need to confirm solid silver and where to find it?
- You would need the density value for pure silver from reference data tables (e.g., CRC Handbook, NIST WebBook, or standard chemistry text) to compare with your measured density
- If the material’s density matches the standard density of silver, and the object has no porosity or hollow regions, it supports it being solid silver; to prove composition beyond doubt, analytical techniques (e.g., spectroscopy, X-ray diffraction) could be used

Given:
- Mass: m = 28.0\ \mathrm{g}
- Density: \rho = 0.7781\ \mathrm{g\,mL^{-1}}
Steps:
- Volume: V = \frac{m}{\rho} = \frac{28.0}{0.7781} \approx 35.98\ \mathrm{mL}
- Significant figures: input masses/densities imply 3 s.f; round to 3 s.f → V \approx 36.0\ \mathrm{mL}
Answer: 36.0\ \mathrm{mL}

Given:
- Dimensions: 2.5\ \mathrm{cm} \times 1.25\ \mathrm{cm} \times 5.75\ \mathrm{cm}
- Mass: m = 132.1\ \mathrm{g}
Steps:
- Volume:
  V = 2.5 \times 1.25 \times 5.75 = 17.96875\ \mathrm{cm^{3}}
- Density:
  \rho = \frac{m}{V} = \frac{132.1}{17.96875} \approx 7.36\ \mathrm{g\,cm^{-3}}
Answer (3 s.f.): \rho \approx 7.36\ \mathrm{g\,cm^{-3}}

Given:
- Lead density: \rho_{\text{Pb}} = 11.34\ \mathrm{g\,cm^{-3}}
- Mass of cube: m = 175\ \mathrm{g}
Steps:
- Volume from mass and density:
  V = \frac{m}{\rho} = \frac{175}{11.34} \approx 15.423\ \mathrm{cm^{3}}
- Cube side:
  a = \sqrt[3]{V} = \sqrt[3]{15.423} \approx 2.49\ \mathrm{cm}
Answer: a \approx 2.49\ \mathrm{cm}

Given:
- Density of gold: \rho_{\text{Au}} = 19.3\ \mathrm{g\,cm^{-3}}
- Mass: m = 1.00\ \mathrm{lb}
Steps:
- Convert mass to grams: m = 1.00\ \mathrm{lb} \times 453.592\ \mathrm{g\,lb^{-1}} = 453.592\ \mathrm{g}
- Volume: V = \frac{m}{\rho} = \frac{453.592}{19.3} \approx 23.50\ \mathrm{cm^{3}}
Answer (3 s.f.): V \approx 23.5\ \mathrm{cm^{3}}

Given:
- Zinc density: \rho_{\text{Zn}} = 7.14\ \mathrm{g\,cm^{-3}}
- Water displacement: initial volume V{i} = 162.5\ \mathrm{mL}, final volume V{f} = 186.0\ \mathrm{mL}
Steps:
- Displaced volume:
  V = V{f} - V{i} = 186.0 - 162.5 = 23.5\ \mathrm{mL}
- Mass of zinc:
  m = \rho V = 7.14 \times 23.5 \approx 167.79\ \mathrm{g}
Answer (3 s.f.): m \approx 168\ \mathrm{g}

Given:
- Volume: V = 250.\ \mathrm{mL}
- Density: \rho = 0.902\ \mathrm{g\,mL^{-1}}
Steps:
- Mass: m = \rho V = 0.902 \times 250.0 = 225.5\ \mathrm{g}
Answer (3 s.f.): m \approx 226\ \mathrm{g}

Given:
- Side length: a = 7.5\ \mathrm{cm}
- Mass: m = 50\ \mathrm{g}
Steps:
- Volume of cube:
  V = a^3 = (7.5)^3 = 421.875\ \mathrm{cm^{3}}
- Density:
  \rho = \frac{m}{V} = \frac{50}{421.875} \approx 0.1186\ \mathrm{g\,cm^{-3}}
Answer (3 s.f.): \rho \approx 0.119\ \mathrm{g\,cm^{-3}}

Always identify what is known and what needs to be found before plugging numbers into formulas
For irregular objects, use water displacement to find volume; ensure volume change is read to the appropriate precision
When comparing measured density to a known value, consider measurement uncertainty and significant figures
If densities are provided in g/mL, they are numerically equal to g/cm$^3$; switching between these units is common in problems
In problems involving mass and volume, keep track of units carefully (g, mL, cm$^3$, etc.)
If asked whether an object is a particular pure element, compare the calculated density to the standard density of that element at the given temperature; exact identification may require more advanced techniques beyond density