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Chapter 9 Solutions

Chapter 9 Solutions. 9.7 Molarity. Molarity (M). Molarity is a concentration unit for the moles of solute in the liters (L) of solution. Molarity (M) = moles of solute = moles liter of solution L Examples:

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Chapter 9 Solutions

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  1. Chapter 9 Solutions 9.7 Molarity

  2. Molarity (M) • Molarity is a concentration unit for the moles of solute in the liters (L) of solution. Molarity (M) = moles of solute = moles liter of solution L Examples: 2.0 M HCl = 2.0 moles HCl 1 L 6.0 M HCl = 6.0 moles HCl 1 L

  3. Preparing a 1.0 Molar Solution • A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution.

  4. Calculation of Molarity What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution ? 1. Determine the moles of solute. 4.0 g NaOH x 1 mole NaOH = 0.10 mole 40.0 g NaOH 2. Calculate molarity. 0.10 mole = 0.20 mole = 0.20 M NaOH 0.50 L 1 L

  5. Learning Check Calculate the molarity of an NaHCO3 solution prepared by dissolving 36 g of solid NaHCO3 in water to give a solution volume of 240 mL. 1) 0.43 M 2) 1.8 M 3) 15 M

  6. Solution 2) 1.8 M 36 g x 1 mole NaHCO3 = 0.43 mole NaHCO384 g 0.43 mole NaHCO3 = 1.8 M NaHCO3 0.240 L

  7. Learning Check A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M

  8. Solution 1) 0.20 M 72 g x 1 mole x 1 = 0.20 moles 180. g 2.0 L 1 L = 0.20 M

  9. Molarity Conversion Factors The units in molarity can be used to write conversion factors.

  10. Learning Check Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 mL of stomach acid? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 mole HCl

  11. Solution 3) 0.15 mole HCl 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L Molarity factor

  12. Learning Check Calculate the grams of KCl that must be dissolved in water to prepare 0.25 L of a 2.0 M KCl solution. 1) 150 g KCl 2) 37 g KCl 3) 19 g KCl

  13. Solution 3) 37 g KCl Determine the number of moles of KCl. 0.25 L x 2.0 mole KCl = 0.50 moles KCl 1 L Convert the moles to grams of KCl. 0.50 moles KCl x 74.6 g KCl = 37 g KCl 1 mole KCl molar mass of KCl

  14. Learning Check How many milliliters of 6.0 M HNO3 contain 0.15 mole of HNO3? 1) 25 mL 2) 90 mL 3) 400 mL

  15. Solution 1) 25 mL 0.15 mole HNO3 x 1 L x 1000 mL 6.0 moles HNO3 1 L Molarity factor inverted = 25 mL HNO3

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