 # 1 1.11 Density Chapter 1 Matter, Measurements, & Calculations Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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1 1.11 Density Chapter 1 Matter, Measurements, & Calculations Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 Density Compares the mass of an object to its volume. Is the mass of a substance divided by its volume. Density expression Density = mass = g or g = g/cm 3 volume mL cm 3 Note: 1 mL = 1 cm 3 Density

3 Densities of Common Substances

4 Osmium is a very dense metal. What is its density in g/cm 3 if 50.0 g of osmium has a volume of 2.22 cm 3 ? 1) 2.25 g/cm 3 2) 22.5 g/cm 3 3) 111 g/cm 3 Learning Check

5 Given: mass = 50.0 g volume = 22.2 cm 3 Plan: Place the mass and volume of the osmium metal in the density expression. D = mass = 50.0 g volume2.22 cm 3 calculator = 22.522522 g/cm 3 final answer (2) = 22.5 g/cm 3 Solution

6 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 cm 3 Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

7 Density Using Volume Displacement The density of the zinc object is then calculated from its mass and volume. mass = 68.60 g = 7.2 g/cm 3 volume 9.5 cm 3 Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

8 What is the density (g/cm 3 ) 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/cm 3 2) 6.0 g/cm 3 3) 380 g/cm 3 Learning Check object 33.0 mL 25.0 mL

9 Solution Given: 48.0 g Volume of water = 25.0 mL Volume of water + metal = 33.0 mL Need: Density (g/mL) Plan: Calculate the volume difference. Change to cm 3, and place in density expression. 33.0 mL - 25.0 mL= 8.0 mL 8.0 mL x 1 cm 3 = 8.0 cm 3 1 mL Set up Problem: Density = 48.0 g = 6.0 g = 6.0 g/cm 3 8.0 cm 3 1 cm 3

10 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

11 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

12 1) vegetable oil 0.91 g/mL water 1.0 g/mL Karo syrup 1.4 g/mL K W V Solution

13 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 factors1 mL 3.8 g Density as a Conversion Factor

14 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

15 Solution 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 Setup: 875 mL x 0.702 g x 1 kg = 0.614 kg 1 mL 1000 g density metric factor factor

16 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

17 Solution 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 and 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 factorfactor

18 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/cm 3 ) are obtained from the cans? 1) 1.0 L2) 2.0 L3) 4.0 L Learning Check

19 Solution 1) 1.0 L 125 cans x 1.0 lb x 454 g x 1 cm 3 x 1 mL x 1 L 21 cans 1 lb 2.70 g 1 cm 3 1000 mL = 1.0 L

20 Which of the following samples of metals will displace the greatest volume of water? 1 2 3 25 g of aluminum 2.70 g/mL 45 g of gold 19.3 g/mL 75 g of lead 11.3 g/mL Learning Check

21 Solution 25 g of aluminum 2.70 g/mL 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 aluminum 2.70 g 2) 45 g x 1 mL = 2.3 mL gold 19.3 g 3) 75 g x 1 mL = 6.6 mL lead 11.3 g