2 8 density

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2.8

Density

Chapter 2 Measurements

Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Objects that sink in water are more dense than water; objects that float in water are less dense.

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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 expression:

D = mass = g or g = g/cm3 volume mL cm3

Note: 1 mL = 1 cm3

Density


Guide to Calculating Density

3 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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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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STEP 1 Given 50.0 g; 22.2 cm3

Need density, in g/cm3

STEP 2 Plan Write the density expression. D = mass

volumeSTEP 3 Express mass in grams and volume in

cm3

mass = 50.0 g volume = 22.2 cm3

STEP 4 Set up problem D = 50.0 g = 22.522522 g/cm3

2.22 cm3

= 22.5 g/cm3 (3 SF)

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

The density of a solid is determined from its mass and the volume it displaces when submerged in water.


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Density Using Volume Displacement

The density of the object iscalculated from its mass andvolume. mass = 68.60 g = 7.2 g/cm3

volume 9.5 cm3


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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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STEP 1 Given 48.0 g volume of water = 25.0 mL volume of water + metal = 33.0

mL Need Density (g/mL)

STEP 2 Plan Calculate the volume difference, change unit to cm3, and place in density expression.

STEP 3 Express mass in grams and volume in cm3

volume of solid = 33.0 mL - 25.0 mL = 8.0 mL

8.0 mL x 1 cm3 = 8.0 cm3

1 mL

STEP 4 Set 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.


Objects that sink in water are more dense than water.

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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

W

W

K

K

K

V

V

V

W

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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For a density of 3.8 g/mL, • an equality is written as

3.8 g = 1 mL

• and two conversion factors are written as

Conversion 3.8 g and 1 mL factors 1 mL 3.8 g

Density as a Conversion Factor


Guide to Using Density

15 Basic Chemistry Copyright © 2011 Pearson Education, Inc.

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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

STEP 1 Given D = 0.702 g/mL; V= 875 mL

Need mass in kg of octane

STEP 2 Plan mL g kg

STEP 3 Equalities

0.702 g = 1 mL 1 kg = 1000 g

STEP 4 Set Up Problem 875 mL x 0.702 g x 1 kg = 0.614 kg (1)

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

STEP 1 Given D = 0.92 g/mL mass = 285 g

Need volume in liters

STEP 2 Plan g mL L

STEP 3 Equalities 1 mL = 0.92 g 1 L = 1000 mL

STEP 4 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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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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Plan: Calculate the volume for each metal and select the metal sample with the greatest volume.

1) 25 g 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
