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Entry Task: April 25 th -26 th Block 2

Entry Task: April 25 th -26 th Block 2. Question- What is the difference between solute and solvent? You have ~ 5 minutes to answer. Agenda:. Sign off and discuss Ch. 15.1 notes Backside of 15.1 notes will be Ch. 15.2 notes HW: Concentration ws . Solutions.

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Entry Task: April 25 th -26 th Block 2

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  1. Entry Task: April 25th-26th Block 2 Question- What is the difference between solute and solvent? You have ~5 minutes to answer

  2. Agenda: • Sign off and discuss Ch. 15.1 notes • Backside of 15.1 notes will be Ch. 15.2 notes • HW: Concentration ws.

  3. Solutions The two components in a solution are: Solvent Larger portion Liquid Does the dissolving Solution Solute Smaller portion Solid Being dissolved

  4. Identify the solute and Solvent for the following

  5. 1) Water 2) salt Solute Solvent

  6. 1) CO2 2) Sugar water Solute Solvent

  7. 1) N2 2) O2 Solute Solvent

  8. Soluble vs. Insoluble • Soluble (miscible) - solute will dissolve ex. sugar in water • Insoluble (immiscible)- solute will not dissolve ex. oil and water “Like dissolves like”

  9. Solvation: • The solvent particles will pull the solute particles apart and surround them. http://programs.northlandcollege.edu/biology/Biology1111/animations/dissolve.html

  10. http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.htmlhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html

  11. What is the general rule to solvation? And what does it mean? • “Like dissolves like”. • If the solute is polar it will dissolve in polar solvents. • If the solute is nonpolar it will dissolve in nonpolar solvents.

  12. As sodium chloride dissolves in water, what happens to the sodium and chloride ions? Look at Fig. 15-3 • Sodium ions and chloride ions are being pulled away from each other by the water molecules

  13. Explain the orientation of the water molecules around the sodium ions and chloride ions. Look at Fig. 15-3 • Sodium ions (Na+) are being pulled away by the negative ends of the water. • Chloride ions (Cl-) are being pulled away by the positive ends of the water.

  14. How does the strength of the attraction between water molecules and sodium and chloride ions compare with the strength of the attraction between the sodium ions and chloride ions? How do you know? • The strength of the water molecules to the ions is greater than the attractive forces between the ions themselves. • IF the attraction (ion to ion) was greater than that of water, it would not have dissolved.

  15. Factors Affecting Solvation: • Agitating the mixture- moves particles through the water to increase more collisions • Increase surface area of the solute. Greater the surface area increase for more collisions • Increase the temperature of the solvent. Adds kinetic energy, increase collisions

  16. Solubility • Solubility is the maximum amount of solute a solvent can dissolve at a specified pressure and temperature • Saturated is a maximum amount of a specific solute in 100 grams of water at a specific temperature and pressure. • Unsaturatedis less than the amount of a specific solute in 100 grams of water at a specific temperature and pressure. • Supersaturated is a more than the maximum amount of a specific solute in 100 grams of water at a specific temperature and pressure.

  17. How does pressure and temperature affect the solubility of a gas in solution? • Pressure- increase in pressure forces gases to stay in solution. If you decrease pressure- gas escapes • Temperature- increase in temperature of a gas solution drives the gas to leave (increasing kinetic energy) and decreasing surface tension.

  18. State Henry’s Law. Provide the equation. • It states that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. S1 = S2 P1 P2

  19. From the Practice Problems provide the answers. Pg. 461 • If 0.55g of a gas dissolves in 1.0 L water at 20.0kPa of pressure, how much will dissolve at 110.0 kPa of pressure? S1 = S2 S1= 0.55g P1 P2 P1= 20.0 kPa 0.55g = X S2 = X g 20.0 110.0 P2= 110.0 kPa 60.5 = 3.0 g/L 20

  20. From the Practice Problems provide the answers. Pg. 461 • A gas has a solubility of 0.66g/L at 10.0 atm of pressure. What is the pressure on a 1.0L sample that contains 1.5 g of gas? S1 = S2 S1= 0.66g P1 P2 P1= 10.0 kPa 0.66g = 1.5 g S2 = 1.5 g 10.0 X P2= X kPa 15 = 23 atm 0.66

  21. Solution Concentration Concentration of a solution is measure of how much solute is dissolved in a specific amount of solvent or solution.

  22. Solution Concentration • Measure by: • Percent by mass • Percent by volume • Molarity • Molality • Mole Fraction

  23. Percent by Mass You have 3.6g NaCl dissolved in 100.0g of H2O. What is the percent by mass of the solution? 3.6 g NaCl + 3.6 g NaCl = 103.6g 100 g H2O 3.4 % NaCl solution 3.6 g NaCl X 100 103.6 g solution

  24. Percent by Mass How much BaCl was added to 325 mls of solution to yield a 23% concentration? = 23% concentration X g NaCl 325 ml solution X g BaCl = 23 325 ml solution 100 = 7475 100 74.5 g BaCl

  25. Percent by Mass How much BaCl was added to 325 mls of solution to yield a 23% concentration? = 23% concentration X g NaCl 325 ml solution HOW MUCH WATER WOULD BE IN THIS SOLUTION? 74.5 g BaCl - 325 ml of solution 250.5 ml of solvent

  26. Percent by Mass A 45% salt solution was prepare by adding 30 grams of salt to an unknown amount of water. How much water is needed to complete this solution? = 45% concentration 30 g NaCl X amount of water + 30 g of solute 30 g NaCl 45(X + 30) = 3000 = 45 X + 30 100 45X + 1350 = 3000 -1350 = 3000 - 1350 45X = 1650 X = 1650 45 X = 36.6ml of water

  27. Percent by Mass A 45% salt solution was prepare by adding 30 grams of salt to an unknown amount of water. How much water is needed to complete this solution? = 45% concentration 30 g NaCl X amount of water + 30 g of solute 30 g NaCl = 45 X + 30 100 74.5 g BaCl - 325 ml of solution 250.5 ml of solvent

  28. Percent by Volume What is the percent by volume of ethanol in a solution that contains 35 ml of ethanol dissolved in 115 ml of water? 35 ml Ethanol 115 water + 35 ml of ethanol = 150 ml solution 23.3 % Ethanol solution 35 ml Ethanol X 100 150 ml solution

  29. Percent by Volume If you have 100 mls of a 30.0% aqueous solution of ethanol, what volumes of ethanol and water are in the solution? 30% or 30 mls is ethanol 70 mls is water

  30. Percent by Volume What is the percent by volume of isopropyl alcohol in a solution that contains 24mL of isopropyl alcohol in 1.1 L of water 24 ml isopropyl alcohol 1.1L or 1100 ml water + 24 ml of ethanol = 1124 ml solution 2.1 % solution 24 ml Ethanol X 100 1124 ml solution

  31. Molarity Calculate the molarity of 2.5 mol of KCl in 2.0L of solution? 2.5 mole KCl = 1.3 M 2.0 L solution

  32. Molarity Calculate the molarity of 1.60L of solution containing 1.55g of dissolved KBr (MM = 119g)? 1.55 g of KBr 1 mole KBr = 0.013 mole KBr 119 g KBr 0.013 mole KBr = 0.008 M 1.60 L solution

  33. Molarity What is the molarity of a bleach solution containing 9.5 g of NaOCl per liter of bleach? 9.5 g of NaOCl 1 mole of NaOCl 74.44 g NaOCl 0.128 moles of NaOCl = 0.128 M 1.0 L solution

  34. Molarity How many moles of CaCl2 are in a 500 ml (0.50 L) 0.128M solution? X moles of CaCl2 = 0.128 M 0.50 L solution (0.50)(0.128) = 0.064 mol CaCl2

  35. Molarity How many grams of CaCl2 are in a 750 ml (0.750 L) 2.10 M solution? X moles of CaCl2 = 2.10M 0.75 L solution (0.75)(2.10) = 1.58 mol CaCl2 111 g CaCl2 1 moles of CaCl2 175 g CaCl2

  36. HW: Concentration ws

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