Question

density practice
$\boldsymbol{d = \frac{mass}{volume}}$
solve the following problems. show all of your work!

1. what is the density of an object with a mass of 45.8g and a volume 12.998 ml?
2. what is the mass of an object with a density of 0.356 g/ml and volume of 125 ml?
3. isopropyl alcohol has a density of 0.785 g/ml. what volume should be measured to obtain 20.0 g of the liquid?
4. what is the density of a 0.741 g cube of aluminum when the cube measures 0.65 cm on a side? (hint: volume using the equation l × w × h.)
5. a graduated cylinder contains 20.0 ml of water. a sample of 125.44 g of copper metal is dropped into the cylinder, raising the level of water in it to 34.0 ml. what is the density of the copper sample?
6. a graduated cylinder contains 10.0 ml of water. a sample of 393.7 g of iron metal is dropped into the cylinder, raising the level of water in it to 60.0 ml. what is the density of the iron sample?
7. a graduated cylinder contains 50.0 ml of water. a rubber stopper with a mass of 9.1 g is dropped into the cylinder, raising the level of water in it to 56.0 ml. what is the density of the rubber stopper?

Explanation:

Step1: Substitute values into density formula

$D = \frac{45.8\ \text{g}}{12.998\ \text{mL}} \approx 3.52\ \text{g/mL}$

Step2: Rearrange density formula for mass

$\text{mass} = D \times \text{volume} = 0.356\ \text{g/mL} \times 125\ \text{mL} = 44.5\ \text{g}$

Step3: Rearrange density formula for volume

$\text{volume} = \frac{\text{mass}}{D} = \frac{20.0\ \text{g}}{0.785\ \text{g/mL}} \approx 25.5\ \text{mL}$

Step4: Calculate cube volume first

$\text{volume} = (0.65\ \text{cm})^3 = 0.274625\ \text{cm}^3$
$D = \frac{0.741\ \text{g}}{0.274625\ \text{cm}^3} \approx 2.7\ \text{g/cm}^3$

Step5: Find copper volume via water displacement

$\text{volume} = 34.0\ \text{mL} - 20.0\ \text{mL} = 14.0\ \text{mL}$
$D = \frac{125.44\ \text{g}}{14.0\ \text{mL}} \approx 8.96\ \text{g/mL}$

Step6: Find iron volume via water displacement

$\text{volume} = 60.0\ \text{mL} - 10.0\ \text{mL} = 50.0\ \text{mL}$
$D = \frac{393.7\ \text{g}}{50.0\ \text{mL}} = 7.87\ \text{g/mL}$

Step7: Find stopper volume via water displacement

$\text{volume} = 56.0\ \text{mL} - 50.0\ \text{mL} = 6.0\ \text{mL}$
$D = \frac{9.1\ \text{g}}{6.0\ \text{mL}} \approx 1.5\ \text{g/mL}$

Answer:

1. $3.52\ \text{g/mL}$
2. $44.5\ \text{g}$
3. $25.5\ \text{mL}$
4. $2.7\ \text{g/cm}^3$
5. $8.96\ \text{g/mL}$
6. $7.87\ \text{g/mL}$
7. $1.5\ \text{g/mL}$