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Section 6.2—Concentration

Section 6.2—Concentration. How do we indicate how much of the electrolytes are in the drink?. 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.

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Section 6.2—Concentration

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  1. Section 6.2—Concentration How do we indicate how much of the electrolytes are in the drink?

  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 gives the ratio of amount dissolved to total amount • There are several ways to show concentration

  4. Percent Weight/Volume • This is a method of showing concentration that is not used as often in chemistry • However, it’s used often in the food and drink industry • For example, your diet drink can might say you have less than 0.035 g of salt in 240 mL. • That would give you a concentration of 0.035 g / 240 mL, which is 0.015% solution

  5. %(W/V) Example Example: If you dissolve 12 g of sugar in 150 mL of water, what percent weight/volume is the solution?

  6. %(W/V) Example Example: If you dissolve 12 g of sugar in 150 mL of water, what percent weight/volume is the solution? 8.0% (W/V)

  7. %(W/V) Example #2 Example: You want to make 200 mL of a 15% (W/V) solution of sugar. What mass of sugar do you need to add to the water?

  8. %(W/V) Example #2 Example: You want to make 200 mL of a 15% (W/V) solution of sugar. What mass of sugar do you need to add to the water? 30 g of sugar

  9. Concentration using # of molecules • When working with chemistry and molecules, it’s more convenient to have a concentration that represents the number of molecules of solute rather than the mass (since they all have different masses) • Remember, we use moles as a way of counting molecules in large numbers

  10. Quick Mole Review (Remember the mole road map ?!?) • 1 mole = 6.02 × 1023 molecules • The molecular mass of a molecule is found by adding up all the atomic masses in the atom • Molecular mass in grams = 1 mole of that molecule

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

  12. 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 0.44 58.44 g NaCl

  13. Molarity • Molarity (M) is a concentration unit that uses moles of the solute instead of the mass of the solute • Molarity Simulation

  14. Molarity Example Example: If you dissolve 12 g of NaCl in 150 mL of water, what is the molarity?

  15. 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

  16. Converting between the two • If you know the %(W/V), you know the mass of the solute • You can convert that mass into moles using molecular mass • You can then use the moles solute to find molarity

  17. Converting from % to M Example Example: What molarity is a 250 mL sample of 7.0 %(W/V) NaCl?

  18. Na 1  22.99 g/mole 22.99 g/mole =  + 35.45 g/mole Cl 1 35.45 g/mole = 58.44 g/mole Converting from % to M Example Example: What molarity is a 250 mL sample of 7.0 %(W/V) NaCl? ? = 17.5 g NaCl 1 mole NaCl molecules = 58.44 g 17.5 g NaCl mole NaCl 1 = _______ mole NaCl 0.30 58.44 g NaCl Remember to change mL to L! 250 mL of water = 0.250 L 1.2 M NaCl

  19. Let’s Practice #2 Example: What is the %(W/V) of a 500. mL sample of a 0.25 M CaCl2 solution?

  20. Ca 1  40.08 g/mole 40.08 g/mole =  + 70.90 g/mole Cl 2 35.45 g/mole = 110.98 g/mole Let’s Practice #2 Example: What is the %(W/V) of a 500. mL sample of a 0.25 M CaCl2 solution? ? = 0.125 moles CaCl2 1 mole CaCl2 molecules = 110.98 g 0.125 moles CaCl2 g CaCl2 110.98 = _______ g CaCl2 13.9 1 mole CaCl2 2.8 %(W/V) CaCl2

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

  22. Concentration of Electrolytes • An electrolyte breaks up into ions when dissolved in water • You have to take into account how the compound breaks up to determine the concentration of the ions CaCl2 Ca+2 + 2 Cl-1 For every 1 CaCl2 unit that dissolves, you will produce 1 Ca+2 ion and 2 Cl-1 ions If the concentration of CaCl2 is 0.25 M, the concentration of Ca+2 is 0.25 M and Cl-1 is 0.50 M

  23. 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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