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Unit 4: Solutions and Solubility

Unit 4: Solutions and Solubility. 6.3 Solution Concentration. What is “ Solution Concentration ” ?. Concentration = the quantity of a given solute in a solution In general, Concentration = quantity of solute quantity of solution. Some Important Terms.

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Unit 4: Solutions and Solubility

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  1. Unit 4: Solutions and Solubility 6.3 Solution Concentration

  2. What is “Solution Concentration”? • Concentration = the quantity of a given solute in a solution • In general, • Concentration = quantity of solute quantity of solution

  3. Some Important Terms • Dilute = having a relatively small quantity of solute per unit volume of solution • Concentrated = having a relatively large quantity of solute per unit volume of solution

  4. Concentration Units

  5. % W/V (%m/v) • This concentration usually describes a solid solute dissolved in a liquid solvent. • Example of a solution that uses this concentration unit: • Intravenous solution is 0.90% (m/v) NaCl (this means 0.90g NaCl dissolved in 100mL of solution)

  6. Example 1 • A box of apple juice has a fructose (sugar) concentration of 12 g/100 mL (12% W/V). What mass of fructose is present in a 175 mL glass of juice)? • volume of solution (apple juice) = 175 mL % W/V = 12 % • mass of solute (fructose) = ? • %W/V = mass of solute X 100 volume of solution mass of solute = % W/V X volume of solution 100 = 12 g/mL X 175 mL 100 = 21 g

  7. Example 1 • A box of apple juice has a fructose (sugar) concentration of 12 g/100 mL (12% W/V). What mass of fructose is present in a 175 mL glass of juice)? • OR • Mass of fructose = 12 g X 175 mL 100 mL = 21 g

  8. % W/W (%m/m) • This concentration usually describes a solid solute in a solid solvent. • Example of a solution that uses this concentration unit: • Toothpaste is 0.24% (m/m) SnF2 (this means 0.24g SnF2 dissolved in100g of solution)

  9. Example 2 • A sterling silver ring has a mass of 12.0 g and contains 11.1 g of pure silver. What is the percentage weight by weight concentration of silver in the metal? • mass of solute = 11.1 g mass of solution = 12.0 g • % W/W = ? • % W/W = mass of solute X 100 mass of solution = 11.1 g X 100 12.0 g = 92.5 %

  10. % V/V (%v/v) • This concentration usually describes a liquid solute in a liquid solvent. • Example of a solution that uses this concentration unit: • Wine is 11.0% (v/v) Ethanol (C2H5OH) (this means 11.0 mL of ethanol is dissolved in 100mL of solution, ie the wine.) % v/v = volume of solute x 100% volume of solution

  11. Example 3 • Gasohol, which is a solution of ethanol and gasoline, is considered to be a cleaner fuel than just gasoline alone. A typical gasohol mixture available across Canada contains 4.1L of ethanol in a 55L tank of fuel. Calculate the percentage by volume concentration of ethanol. • % v/v = volume of ethanol X 100 volume of solution =4.1L X 100 55L = 7.5%

  12. parts per million - ppm • This concentration unit is for very small quantities of solute • c (ppm) = amount of solute amount of solution • Example of a solution that uses this concentration unit: • Bottled water contains 280 ppm of HCO31-, 118 ppm Ca… (this means every 1 000 000 parts of solution contain 280 parts of HCO31- solute) *1ppm = 1 drop of water in a bathtub

  13. Units for ppm 1ppm = 1 mg/L 1 mg/kg 1 μg/g *Choose the unit that matches the information given in the example you are calculating.

  14. Example 4: • If the concentration of oxygen in water is 8 ppm, what mass of oxygen is present in 150 mL of water? c = 8 ppm (mg/L) v (volume of solution) = 150 mL = 0.15 L m (mass of oxygen) = ? c = m v m = c X v m = 8 mg/L X 0.15 L m = 1 mg

  15. Molar Concentration - Molarity (M) • This concentration unit describes the amount of solute (in moles) dissolved in 1 L of solution. • Example of a solution that uses this concentration unit: • Hydrochloric acid - HCl 1.25M (mol/L) This means that 1.25 moles of HCl are dissolved in 1.00L of solution. Chemists use Molarity because it allows them to relate concentration to moles of a substance. Molar concentration = amount of solute (mol) volume of solution (L) C = n v

  16. Example 5: • When 2.00 g of KMnO4 is dissolved into 100.0 mL of solution, what molar concentration results? m = 2.00 g V = 100.0 mL C = ? C = n V n = m M n = 2.00 g 158.04g/mol = 0.0127 mol C = 0.0127 mol 0.1000 L = 0.127 mol/L

  17. Example 6: • How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 mol/L solution? C = 0.200 mol/L V = 500.0 mL (0.5000L) m = ? C = n V n = C X V = 0.200 mol/L X 0.5000L = 0.100 mol m = M X n = 158.04 g/mol X 0.100 mol = 15.8 g

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