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Section 6.2—Concentration

Section 6.2—Concentration. Concentrated versus Dilute. solvent. solute. Lower concentration Not as many solute (what’s being dissolved) particles. Higher concentration More solute (what’s being dissolved) particles. Concentration.

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Section 6.2—Concentration

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  1. Section 6.2—Concentration

  2. Concentrated versus Dilute solvent solute Lower concentration Not as many solute (what’s being dissolved) particles Higher concentration More solute (what’s being dissolved) particles

  3. Concentration • Concentration – a ratio of the amount of solute dissolved over the total amount of solution. • There are several ways to show concentration – we will only focus on one

  4. Quick Mole Review • 1 mole = 6.02 × 1023 molecules • The molar mass of a molecule is found by adding up all the atomic masses in the atom ( from the periodic table) • Molecular mass in grams = 1 mole of that molecule

  5. Quick Mole Example Example: How many moles are in 25.5 g NaCl?

  6. Na 1  22.99 g/mole 22.99 g/mole =  + 35.45 g/mole Cl 1 35.45 g/mole = 58.44 g/mole Quick Mole Example Example: How many moles are in 25.5 g NaCl? 1 mole NaCl molecules = 58.44 g 25.5 g NaCl mole NaCl 1 = _______ mole NaCl 0.436 58.44 g NaCl

  7. Molarity • Molarity (M) is a concentration unit that uses moles of the solute over the total volume of the solution

  8. Molarity Example Example: If you dissolve 12 g of NaCl to make 150 mL of solution, what is the molarity?

  9. Na 1  22.99 g/mole 22.99 g/mole =  + 35.45 g/mole Cl 1 35.45 g/mole = 58.44 g/mole Molarity Example Example: If you dissolve 12 g of NaCl in 150 mL of water, what is the molarity? 1 mole NaCl molecules = 58.44 g 12 g NaCl mole NaCl 1 = _______ mole NaCl 0.21 58.44 g NaCl Remember to change mL to L! 150 mL of water = 0.150 L 1.4 M NaCl

  10. Let’s Practice #3 Example: What are the molarities of the ions made in a 0.75 M solution of Ca(NO3)2

  11. Let’s Practice #3 Example: What are the molarities of the ions made in a 0.75 M solution of Ca(NO3)2 Ca(NO3)2 Ca+2 + 2 NO3-1 For every 1 Ca(NO3)2, there will be 1 Ca+2 and 2 NO3-1 ions Ca+2 = 0.75 M NO3-1 = 1.5 M

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