Question

ws 1.4 density problems (be certain to specify final units)

1. calculate the density of an object which has a volume of 12.5 cm³ and a mass of 19.0 grams. (1.52)
2. an object with a mass of 5.1 grams is placed into a graduated cylinder which reads 10.5 ml. the water subsequently rises to 18.3 ml. what is the density of this object? (0.65)
3. determine the volume of an object which has a density of 0.950 g/ml and a mass of 125 grams. (132)
4. if the density of copper at 20°c is 8.933 g/cm³. if the mass of a penny is 3.16 grams, what would be the volume of a one-cent copper coin at 20°c? (0.354)
5. the density of a sodium hydroxide solution is 1.1589 g/ml. what is the mass of 550.0 ml of this solution? (637.4)
6. what is the volume of a tank that can hold 18.754 kg of methanol whose density is 0.788 g/ml? (2.38×10⁷)
7. what is the density, in g/cm³, of a board whose dimensions are 5.54 cm × 10.6 cm × 199 cm and whose mass is 28.6 kg? (2.45)
8. cacl₂ is used as a de-icer on roads in the winter. it has a density of 2.50 g/cm³. what is the mass of 15.0 l of this substance? (3.75×10⁴)

Explanation:

Step1: Use density formula \(

ho = \frac{m}{V} \)
\(
ho = \frac{19.0\ \text{g}}{12.5\ \text{cm}^3} = 1.52\ \text{g/cm}^3 \)

Step2: Calculate volume (water displacement)

\( V = 18.3\ \text{mL} - 10.5\ \text{mL} = 7.8\ \text{mL} \); \(
ho = \frac{5.1\ \text{g}}{7.8\ \text{mL}} \approx 0.65\ \text{g/mL} \)

Step3: Use volume formula \( V = \frac{m}{

ho} \)
\( V = \frac{125\ \text{g}}{0.950\ \text{g/mL}} \approx 131.6\ \text{mL} \)

Step4: Apply volume formula for copper

\( V = \frac{3.16\ \text{g}}{8.933\ \text{g/cm}^3} \approx 0.354\ \text{cm}^3 \)

Step5: Use mass formula \( m =

ho V \)
\( m = 1.1589\ \text{g/mL} \times 550.0\ \text{mL} = 637.4\ \text{g} \)

Step6: Convert mass to grams, compute volume

\( m = 18.754\ \text{Kg} = 18754\ \text{g} \); \( V = \frac{18754\ \text{g}}{0.788\ \text{g/mL}} \approx 2.38 \times 10^4\ \text{mL} \)

Step7: Calculate volume (dimensions), then density

\( V = 5.54\ \text{cm} \times 10.6\ \text{cm} \times 199\ \text{cm} \approx 11686\ \text{cm}^3 \); \( m = 28.6\ \text{Kg} = 28600\ \text{g} \); \(
ho = \frac{28600\ \text{g}}{11686\ \text{cm}^3} \approx 2.45\ \text{g/cm}^3 \)

Step8: Convert volume, compute mass

\( V = 15.0\ \text{L} = 15000\ \text{mL} \); \( m = 2.50\ \text{g/mL} \times 15000\ \text{mL} = 3.75 \times 10^4\ \text{g} \)

Answer:

1. 1.52 g/cm³
2. 0.65 g/mL
3. 131.6 mL
4. 0.354 cm³
5. 637.4 g
6. 2.38×10⁴ mL
7. 2.45 g/cm³
8. 3.75×10⁴ g