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2.8

Density

Chapter 2 Measurements

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

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Density

• Compares the mass of an object to its volume

• Is the mass of a substance divided by its volume

Density expressionD = mass = g or g = g/cm3 volume mL cm3

Note: 1 mL = 1 cm3

Density

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Densities of Common Substances

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Osmium is a very dense metal. What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3?

1) 2.25 g/cm3

2) 22.5 g/cm3

3) 111 g/cm3

Learning Check

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Given: mass = 50.0 g volume = 22.2 cm3

Plan: Place the mass and volume of the osmium metalin the density expression.

D = mass = 50.0 g volume 2.22 cm3

Calculator = 22.522522 g/cm3

Final answer (2) = 22.5 g/cm3

Solution

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Volume by Displacement

• A solid completely submerged in water displaces its own volume of water.

• The volume of the solid is calculated from the volume difference.

45.0 mL - 35.5 mL

= 9.5 mL = 9.5 cm3

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

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Density Using Volume Displacement

The density of the zinc object isthen calculated from its massand volume.

mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

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What is the density (g/cm3) of 48.0 g of a metal if the level of water in a graduated cylinder rises from 25.0 mL to 33.0 mL after the metal is added?

1) 0.17 g/cm3 2) 6.0 g/cm3 3) 380 g/cm3

25.0 mL 33.0 mL

object

Learning Check

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Given: 48.0 g Volume of water = 25.0 mL Volume of water + metal = 33.0

mLNeed: Density (g/mL)

Plan: Calculate the volume difference. Change to cm3,

and place in density expression.

33.0 mL - 25.0 mL = 8.0 mL

8.0 mL x 1 cm3 = 8.0 cm3

1 mLSet Up Problem:

Density = 48.0 g = 6.0 g = 6.0 g/cm3

8.0 cm3 1 cm3

Solution

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Sink or Float

• Ice floats in water because the density of ice is less than the density of water.

• Aluminum sinks because its density is greater than the density of water. Copyright © 2005 by Pearson Education, Inc.

Publishing as Benjamin Cummings

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Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL), vegetable (V) oil (0.91 g/mL,) water (W) (1.0 g/mL)

1 2 3

K

K

W

W

W

V

V

V

K

Learning Check

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1)

Vegetable oil 0.91 g/mL

Water 1.0 g/mL

Karo syrup 1.4 g/mLK

W

V

Solution

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Density can be written as an equality. • For a substance with a density of 3.8 g/mL, the

equality is: 3.8 g = 1 mL

• From this equality, two conversion factors can be written for density.

Conversion 3.8 g and 1 mL factors 1 mL 3.8 g

Density as a Conversion Factor

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The density of octane, a component of gasoline, is 0.702 g/mL. What is the mass, in kg, of 875 mL of octane?

1) 0.614 kg

2) 614 kg

3) 1.25 kg

Learning Check

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1) 0.614 kg

Given: D = 0.702 g/mL V= 875 mL

Unit plan: mL g kg

Equalities: density 0.702 g = 1 mL

and 1 kg = 1000 g

Set Up: 875 mL x 0.702 g x 1 kg = 0.614 kg Problem 1 mL 1000 g

density metric factor factor

Solution

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If olive oil has a density of 0.92 g/mL, how many liters of olive oil are in 285 g of olive oil?

1) 0.26 L

2) 0.31 L

3) 310 L

Learning Check

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2) 0.31 L

Given: D = 0.92 g/mL mass = 285 g

Need: volume in liters

Plan: g mL L

Equalities: 1 mL = 0.92 g 1 L = 1000 mL

Set Up Problem:

285 g x 1 mL x 1 L = 0.31 L 0.92 g 1000 mL density metric

factor factor

Solution

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A group of students collected 125 empty aluminum cans to take to the recycling center. If 21 cans make 1.0 lb aluminum, how many liters of aluminum (D=2.70 g/cm3) are obtained from the cans?

1) 1.0 L 2) 2.0 L 3) 4.0 L

Learning Check

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1) 1.0 L

125 cans x 1.0 lb x 454 g x 1 cm3 x 1 mL x 1 L 21 cans 1 lb 2.70 g 1 cm3 1000

mL

= 1.0 L

Solution

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Which of the following samples of metals will displace the greatest volume of water?

1 2 3

25 g of aluminum2.70 g/mL

45 g of gold19.3 g/mL

75 g of lead11.3 g/mL

Learning Check

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1)

Plan: Calculate the volume for each metal and select the metal sample with the greatest volume.

1) 25g x 1 mL = 9.3 mL aluminum2.70 g

2) 45 g x 1 mL = 2.3 mL gold19.3 g

3) 75 g x 1 mL = 6.6 mL lead11.3 g

Solution

25 g of aluminum2.70 g/mL