2-5 Density Practice Problems Answers

advertisement
Name_________________ Date______________ Period__________
Density Practice Problems KEY
This image will help you in figuring out how to solve density problems:
Simply cover up whichever value you need to calculate and the other two are shown in their proper
placement, be it to multiply or to divide.
For example, cover up the M. This leave you with dV (ignore the fact that it is in the denominator).
Density times volume will give you mass. You can also check it out by way of the units: (g / cm3) x
cm3 cancels out the volume unit leaving grams, the desired unit for mass.
Solve the following problems showing all your work including equations and units.
1) A block of aluminum occupies a volume of 15.0 mL and weighs 40.5 g. What is its density?
D=m/v
D = 40.5 g / 15 ml = 2.7 g/mL
2) Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used
to fill the cylinder weighs 306.0 g. From this information, calculate the density of mercury.
D=m/v
D = 306.0 g / 22.5 mL = 13.6 g/mL
3) What is the mass of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl
alcohol is 0.789 g/mL.
m=DxV
m = 0.789 g/mL x 200.0 mL = 157.8 g
4) A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by
5.5 cm by 4.6 cm. From this data, what is the density of copper?
V=LxWxH
V = (8.4 cm) (5.5 cm) (4.6 cm) = 212.52 cm3
D=m/V
D = 1896 g / 212.52 cm3 = 8.9 g/cm3
5) 28.5 g of iron shot is added to a graduated cylinder containing 45.50 mL of water. The water level
rises to the 49.10 mL mark. From this information, calculate the density of iron.
Volume of iron shot = 49.10 mL - 45.50 mL = 3.60 mL
D=m/V
D = 28.5 g / 3.60 mL = 7.92 g/mL
6) An irregular object with a mass of 18 kg displaces 2.5 L of water when placed in a large overflow
container. Calculate the density of the object.
1 kg = 1000 g
&
1 L = 1000 mL
18 kg = 18000 g
&
2.5 L = 2500 mL
D=m/V
D = 18 kg / 2.5 L = 7.2 kg/L or
D = 18000 g / 2500 mL = 7.2 g/mL or 7.2 g/cm3
BONUS: A graduated cylinder has a mass of 80 g when empty. When 20 mL of water is added, the
graduated cylinder has a mass of 100 g. If a stone is added to the graduated cylinder, the water level
rises to 45 mL and the total mass is now 156 g. What is the density of the stone?
Stone mass = 156 g – 100 g = 56 g
Stone volume = 45 mL – 20 mL = 25 mL
D=m/V
D = 56 g / 25 mL = 2.24 g/mL