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Review of Solution Concentration and Molarity in Chemistry

This review focuses on key concepts related to solution concentration, particularly molarity, and how changes in volume affect concentration. When water evaporates from a solution, its molarity increases as the solute becomes more concentrated. The review also includes examples of naturally diluting solutions, explains the concept of dilution, and provides equations for calculating molarity and preparing solutions from concentrated stock. Understanding these principles is crucial for effective solution preparation in laboratory settings.

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Review of Solution Concentration and Molarity in Chemistry

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  1. Unit 10B Review Reg Chem 2012-13

  2. 11 • When a solution sits out over a long period of time and water evaporates the concentration of the solution __________. Increases. As the volume of solution decreases, the molarity (concentration) increases

  3. 12 • List an example of a solution that would naturally dilute. Ice in pop

  4. 13 • What does solution concentration describe? Molarity: amount of solute dissolved in a given amount of solvent

  5. 14 • What do we use to describe solution concentration in chemistry? Molarity

  6. 15 • If a solution is “strong” it is • If a solution is “weak” it is concentrated dilute

  7. 16 • What does it mean to dilute a solution? What equation do we use for dilutions? To lower its concentration “water it down” Yet keep the number of moles the same M1V1 = M2V2

  8. 17 • What is the molarity of a sodium chloride solution that contains 1.73 moles in 3.94 L of solution?

  9. 18 • What is the molarity of sodium hydroxide solution that contains 23.5 g NaOH in 500.0 mL of solution?

  10. 19 • How many grams of potassium nitrate are in 275 mL of 1.25 M solution?

  11. 20 • How many mL of 3.25 M hydrochloric acid would contain 16.0 grams of solute?

  12. 21 • You have 12.0 M HCl in your stock room, how would you prepare 600.0 mL of 2.50 M HCl solution? Measure out 125 mL of stock solution. Add 475 mL of distilled water. (12.0 M)(x) = (2.50 M)(600.0 mL) X=125 mL

  13. 22 • How would you correctly prepare 500.0 mL of a 3.0 M solution of NaOH from solid solute? NaOH Measure out 60 g of NaOH (s) Add water to the 500.0 mL line

  14. 23 • How would you prepare 500 mL of 3.0 M NaOH from 12.0 M concentrated stock solution? Measure out 125 mL of stock solution. Add it to 375 mL of distilled water (12.0 M)(X) = (3.0 M)(500 mL) X=125 mL

  15. 26 • An excess of zinc is added to 125 mL of 0.100 M HCl solution. What mass of zinc chloride is formed? Zn + 2HCl  ZnCl2 + H2

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