1. 1 2.8 Density Chapter 2Measurements Copyright © 2008 by Pearson Ed ucation, Inc. Publishing as Benjamin Cummings

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

3. 3 Densities of Common Substances Copyright © 2008 by Pearson Ed ucation, Inc. Publishing as Benjamin Cummings

4. 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. 5 Given: mass = 50.0 g volume = 22.2 cm 3 Plan: Write the density expression. D = mass volume Express mass in grams and volume in cm 3 mass = 50.0 g volume = 22.2 cm 3 Set up problem using mass and volume. D = 50.0 g = 22.522522 g/cm 3 2.22 cm 3 = 22.5 g/cm 3 (3 SF) Solution

6. 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 © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

7. 7 Density Using Volume Displacement The density of the object is calculated from its mass and volume. mass = 68.60 g = 7.2 g/cm 3 volume 9.5 cm 3 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

8. 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 25.0 mL 33.0 mL object Learning Check

9. 9 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 Solution

10. 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 © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

11. 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. 12 1) vegetable oil 0.91 g/mL water 1.0 g/mL Karo syrup 1.4 g/mL K W V Solution

13. 13 Density can be written as an equality. For a density of 3.8 g/mL, the equality is: 3.8 g = 1 mL From this equality, two conversion factors can be written: Conversion 3.8 g and 1 mL factors1 mL 3.8 g Density as a Conversion Factor

14. 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. 15 1) 0.614 kg Given: D = 0.702 g/mL V= 875 mL Need: mass in kg of octane 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 Solution

16. 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. 17 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 factorfactor Solution

18. 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. 19 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 Solution

20. 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. 21 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 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 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 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