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Friday, Feb. 21 st : “A” Day Monday, Feb. 24 th : “B” Day Agenda

Friday, Feb. 21 st : “A” Day Monday, Feb. 24 th : “B” Day Agenda. Collect chromatography labs Begin Section 13.2: “Concentration and Molarity” Demos: Preparing 250 mL of a 0.5000 M CuSO 4 solution Do solvents always add up? Homework: Practice pg. 461: #2, 3, 5, 6

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Friday, Feb. 21 st : “A” Day Monday, Feb. 24 th : “B” Day Agenda

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  1. Friday, Feb. 21st: “A” DayMonday, Feb. 24th: “B” DayAgenda • Collect chromatography labs • Begin Section 13.2: “Concentration and Molarity” • Demos: • Preparing 250 mL of a 0.5000 M CuSO4 solution • Do solvents always add up? • Homework: • Practice pg. 461: #2, 3, 5, 6 • Practice pg. 465: #1-7

  2. Concentration • In a solution, the solute is distributed evenly throughout the solvent. This means that any part of a solution has the same ratio of solute to solvent as any other part of the solution. • This ratio is the concentration of the solution. • Concentration: the amount of a particular substance in a given quantity of a solution

  3. Calculating Concentration • Concentrations can be expressed in many forms.

  4. Calculating Concentrations • One unit of concentration used in pollution measurements that involve very low concentrations is parts per million,or ppm. • Parts per million is the number of grams of solute in 1 million grams of solution.

  5. Sample Problem A, Pg. 461 A chemical analysis shows that there are 2.2 mg of lead in exactly 500 g of a water sample. Convert this measurement to parts per million. First, change mg g: 2.2 mg Pb X 1 g = .0022 g Pb 1,000 mg Divide by 500 g to get the amount of Pb in 1 g of H2O. Then multiply by 1,000,000 to get the amount of Pb in 1,000,000 g H2O: .0022 g Pb X 1,000,000 parts = 4.4 ppm Pb 500 g 1 million (2 sig figs)

  6. Additional Example How many parts per million of mercury are there in a sample of tap water with a mass of 750 g containing 2.2 mg of Hg? First, change mg g: 2.2 mg Hg X 1 g = .0022 g Hg 1,000 mg Divide by 750 g to get the amount of Hg in 1 g of H2O. Then multiply by 1,000,000 to get the amount of Hg in 1,000,000 g H2O: .0022 g X 1,000,000 parts = 2.9 ppm Hg 750 g 1 million (2 sig figs)

  7. Molarity • Since the mole is the unit chemists use to measure the number of particles, they often specify concentrations using molarity. • Molarity (M): a concentration unit of a solution expressed as moles of solute dissolved per liter of solution. Molarity (M) = moles of solute L of solution

  8. Molarity Example • Suppose that 0.30 moles of KBr are present in 0.40 L of solution. • The molarity of the solution is calculated as follows: 0.30 mol KBr = 0.75 M KBr 0.40 L solution • This is called a 0.75 molar solution of KBr.

  9. Preparing a Solution of a Specified Molarity • Note that molarity describes concentration in terms of volume of solution, NOT volume of solvent. • If you simply added 1.000 mol solute to 1.000 L solvent, the solution would not be 1.000 M. • The added solute would increase the volume, so the solution would not have a concentration of 1.000 M. • The solution must be made to have exactly the specified volume of solution.

  10. Demo: Preparing 250 mL of a 0.5000 M CuSO4 Solution (pg. 463)

  11. Calculating Molarity • In working with solutions in chemistry, you will find that numerical calculations often involve molarity. • The key to all such calculations is the definition of molarity… Molarity (M) = moles of solute L of solution

  12. Calculating Molarity Given Mass of Solute and Volume of Solution

  13. Sample Problem B, Pg. 465 • What is the molarity of a potassium chloride solution that has a volume of 400.00 mL and contains 85.0 g KCl? Molarity = moles of solute L of solution First, use molar mass to change g of KCl → moles KCl: 85.0 g KCl X 1 mol KCl = 1.14 mol KCl 74.6 g KCl 1.14 mol KCl = 2.85 M KCl .400 L (3 sig figs)

  14. Additional Example • Determine the molarity of a solution prepared by dissolving 16.9 g of NaOH in enough water to make 250.00 mL of solution. Molarity = moles of solute L of solution First, use molar mass to change g NaOH mol NaOH 16.9 g NaOH X 1 mol NaOH = 0.423 mol NaOH 40 g NaOH 0.423 mol NaOH = 1.69 M NaOH 0.250 L (3 sig figs)

  15. Calculating Mass of Solute Given Molarity and Volume of Solution

  16. Additional Example • How many grams of NaOH are needed to prepare 250.0 mL of a 1.69 M NaOH solution? First, change mL L: 250 mL X 1 L = 0.2500 L 1,000 mL Then, multiply by molarity to find moles of solute: 0.2500 L X 1.69 moles NaOH = 0.423 mol NaOH 1 L Finally, use molar mass to find mass of solute: 0.423 mol NaOH X 40 g NaOH = 16.9 g NaOH 1 mole NaOH (3 sig figs)

  17. Additional Example • How many grams of glucose, C6H12O6, are in 255 mL of a 3.55 M solution? First, change mL L: 255 mL X 1 L = 0.255 L 1,000 mL Then, multiply by molarity to find moles of solute: 0.255 L X 3.55 moles glucose = 0.905 mol 1 L glucose Finally, use molar mass to find mass of solute: 0.905 mol glucose X 180 g glucose = 163 g glucose 1 mol glucose

  18. Demo: Do Solvents Always Add Up? 50.0 mL H2O + 50.0 mL ethanol = ________?___ mL solution +

  19. Homework • Practice pg. 461: # 2, 3, 5, 6 • Practice pg. 465: # 1-7 We will finish section 13.2 next time…

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