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MOLARITY – Ch 13, p. 412

Variables. Units. MOLARITY – Ch 13, p. 412. Quantifies the concentration of a solution. Molarity (M) = mol solute = n = mol volume solution V L Read as “moles solute per liter of solution”. MOLARITY – Ch 13.

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MOLARITY – Ch 13, p. 412

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  1. Variables Units MOLARITY – Ch 13, p. 412 Quantifies the concentration of a solution. Molarity (M) = mol solute = n = mol volume solution V L Read as “moles solute per liter of solution”

  2. MOLARITY – Ch 13 Example 1: A bottle labeled “0.2 M NaCl” reads as “0.2 molar sodium chloride solution.”

  3. MOLARITY – Ch 13 Example 1: A bottle labeled “0.2 M NaCl” reads as “0.2 molar sodium chloride solution.” This means 1 L of the solution contains 0.2 moles NaCl.

  4. MOLARITY – Ch 13 Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution.

  5. Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given:

  6. Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given: mass = 5.844 g

  7. Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given: mass = 5.844 g V = 200.0 mL

  8. Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given: mass = 5.844 g V = 200.0 mL Unknown:

  9. Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given: mass = 5.844 g V = 200.0 mL Unknown: M

  10. Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given: mass = 5.844 g V = 200.0 mL Unknown: M Need:

  11. MOLARITY – Ch 13 Example 2: Calculate the molarity of a solution prepared by dissolving 5.844 g NaCl in enough water to make a 200.0 mL solution. Given: mass = 5.844 g V = 200.0 mL Unknown: M Need: molar mass NaCl

  12. MOLARITY – Ch 13 Example 2: Given: mass = 5.844 g V = 200.0 mL = 0.2000 L Unknown: M Need: molar mass NaCl 58.443 g/mol M = n = 5.844 g V

  13. MOLARITY – Ch 13 Example 2: Given: mass = 5.844 g V = 200.0 mL = 0.2000 L Unknown: M Need: molar mass NaCl M = n = 5.844 g mol V 58.443 g

  14. MOLARITY – Ch 13 Example 2: Given: mass = 5.844 g V = 200.0 mL = 0.2000 L Unknown: M Need: molar mass NaCl M = n = 5.844 g mol V 58.443 g

  15. MOLARITY – Ch 13 Example 2: Given: mass = 5.844 g V = 200.0 mL = 0.2000 L Unknown: M Need: molar mass NaCl M = n = 5.844 g mol V 0.2000 L 58.443 g

  16. MOLARITY – Ch 13 Example 2: Given: mass = 5.844 g V = 200.0 mL = 0.2000 L Unknown: M Need: molar mass NaCl M = n = 5.844 g mol 0.49997 M V 0.2000 L 58.443 g

  17. MOLARITY – Ch 13 Example 2: Given: mass = 5.844 g V = 200.0 mL = 0.2000 L Unknown: M Need: molar mass NaCl M = n = 5.844 g mol 0.5000 M V 0.2000 L 58.443 g

  18. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH.

  19. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given:

  20. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL

  21. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL molarity = 0.250 M

  22. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH

  23. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH Need:

  24. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH Need: M = n V

  25. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  V

  26. MOLARITY – Ch 13 Example 3: Calculate the moles of NaOH present in 5000.0 mL of 0.250 M NaOH. Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V

  27. MOLARITY – Ch 13 Example 3: Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V 0.250 M

  28. MOLARITY – Ch 13 Example 3: Given: V = 5000.0 mL molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V 0.250 mol L

  29. MOLARITY – Ch 13 Example 3: Given: V = 5000.0 mL = 5.0000 L molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V 0.250 mol L

  30. MOLARITY – Ch 13 Example 3: Given: V = 5000.0 mL = 5.0000 L molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V 0.250 mol 5.0000 L L

  31. MOLARITY – Ch 13 Example 3: Given: V = 5000.0 mL = 5.0000 L molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V 0.250 mol 5.0000 L L

  32. MOLARITY – Ch 13 Example 3: Given: V = 5000.0 mL = 5.0000 L molarity = 0.250 M Unknown: n, mol NaOH Need: M = n  n = M•V V 0.250 mol 5.0000 L 1.25 mol L

  33. MOLARITY – Ch 13 Example 4: Calculate the volume of 0.50 M HCl that would contain 2.0 mol HCl. YOU TRY IT!

  34. MOLARITY – Ch 13 Example 4: Given: n = 2.0 mol HCl molarity = 0.50 M Unknown: V Need: M = n  V = n V M 2.0 mol L 4.0 L 0.50 mol

  35. MOLARITY – Ch 13 Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl? YOU TRY IT!

  36. MOLARITY – Ch 13 Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl? Use C1V1 = C2V2

  37. MOLARITY – Ch 13 Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl? Use C1V1 = C2V2 Given: C1 = 0.50 M V1 = 100mL C2 = 12 M

  38. MOLARITY – Ch 13 Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl? Use C1V1 = C2V2 Given: C1 = 0.50 M V1 = 100mL C2 = 12 Unknown:

  39. MOLARITY – Ch 13 Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl? Use C1V1 = C2V2 Given: C1 = 0.50 M V1 = 100mL C2 = 12 Unknown: V2 =

  40. MOLARITY – Ch 13 Example 5: How would you prepare 100 mL of 0.50 M HCl from concentrated 12 M HCl? Use C1V1 = C2V2  C1V1 = V2 C2 Given: C1 = 0.50 M V1 = 100mL C2 = 12 Unknown: V2 =

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