Density Practice Problems - New Canaan
Density Practice Problems
The density of a substance is a measure of how much mass is packed into a certain volume of the substance. Substances with a high density, like steel, have molecules that are packed together tightly. Substances with a low density, like cork, have fewer molecules packed into the same amount of space.
The density of a substance can be found by dividing its mass by its volume. As long as a substance is homogeneous, the size or shape of the sample doesn’t matter. The density will always be the same. This means that a steel paper clip has the same density as a steel girder used to build a bridge.
Density = Mass ÷ Volume D =
Use the density formula to solve the following problems. Show all work and the answer must have the correct units. Remember that volume can have different forms. A block of ice with a volume of 3 cm3 would be 3 mL of liquid after being melted.
1. What is the density of CO gas if 0.196 g occupies a volume of 100 ml?
m = 0.196 g D = m/v
v = 100 ml D = 0.196 g / 100 ml
D = 0.00196 g/ml
Answer_0.00196 g/ml
2. A block of wood 3 cm on each side has a mass of 27 g. What is the density of the block? (Hint, don’t forget to find the volume of the wood.)
l = 3 cm v = l x w x h
w = 3 cm v = 3 cm x 3 cm x 3cm
h = 3 cm v = 27 cm3
m = 27 g
D = m/v
D = 27 g / 27 cm3
D = 1 g/cm3
Answer 1 g/cm3
3. An irregularly shaped stone was lowered into a graduated cylinder holding a volume of water equal to 2 ml. The height of the water rose to 7 ml. If the mass of the stone was 25 g, what was its density?
vi = 2 ml v = vf – vi D = m / v
vf = 7 ml v = 7ml – 2 ml D = 25 g / 5 ml
m = 25 g v = 5ml D = 5 g/ml
Answer 5 g/ml
4. A 10.0 cm3 sample of copper has a mass of 89.6 g. What is the density of copper?
v = 10.0 cm3 D = m / v
m = 89.6 g D = 89.6 g / 10.0 cm3
D = 8.96 g/cm3
Answer 8.96 g/cm3
5. Silver has a density of 10.5 grams/cm3 and gold has a density of
19.3 g/cm3. Which would have the greater mass, 5cm3 of silver or 5cm3 of gold?
D (silver) = 10.5 g/cm3 D (gold) = 19.3 g/cm3
v (silver) = 5 cm3 v (gold) = 5 cm3
m (silver) = D x v m (gold) = D x v
m (silver) = 10.5 g/cm3 x 5 cm3 m (gold) = 19.3 g/cm3 x 5 cm3
m (silver) = 52.5 g m (gold ) = 96. 5 g
Answer gold
6. Five mL of ethanol has a mass of 3.9 g, and 5.0 mL of benzene has a mass of 44 g. Which liquid is denser?
v (ethanol) = 5 ml v (benzene) = 5 ml
m (ethanol) = 3.9 g m (benzene) = 44 g
D (ethanol) = m / v D (benzene) = m / v
D (ethanol) = 3.9 g / 5 ml D (benzene) = 44 g / 5 ml
D (ethanol) = 0.78 g/ml D (benzene) = 8.8 g/ml
Answer benzene
7. A sample of iron has the same dimensions of 2 cm x 3 cm x 2 cm. If the mass of this rectangular-shaped object is 94 g, what is the density of iron?
l = 2 cm m = 94 g v = l x w x h D = m / v
w = 3 cm v = 2 cm x 3 cm x 2 cm D = 94 g / 12 cm3
h = 2 cm v = 12 cm3 D = 7.83 g/cm3
Answer 7.83 g/cm3
-----------------------
m
V
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
Related searches
- sat math practice problems printable
- significant figures practice problems worksheet
- theoretical yield practice problems answers
- practice problems for significant figures
- synthesis practice problems and answers
- sig fig practice problems worksheet
- density word problems answer key
- density practice problems worksheet answers
- density word problems pdf
- density word problems answers
- density practice problems with answers
- density practice problems worksheet key