1. Solutions The composition of the solvent and the solute determine whether a substance will dissolve. Stirring, temperature, and surface area of dissolving particles determine how fast the substance will dissolve.

2. Heterogeneous = not the same throughout  Suspensions = look uniform when stirred; however, separate when left alone (Italian dressing, OJ w/ pulp)  Colloids = mixture of very tiny particles dispersed in another substance but do not settle out (gelatin, whipped cream, smoke)  Emulsion = immiscible liquids that are spread throughout one another (mayo, cream, butter)

4. Homogeneous = same throughout  Solution = 2 or more substances uniformly spread throughout a single phase  Solute = substance that dissolves in a solution  Solvent = substance that does the dissolving

5. Hot Water Sugar Sugar Water

6. Solubility The quality or condition of being soluble. The amount of solute that can be dissolved in a given amount of solvent at a certain temperature to produce a saturated solution Often expressed in grams of solute per 100g of solvent

7. Miscible – when two liquids completely dissolve in each other forming a homogeneous solution  Water & ethanol Immiscible – liquids that are insoluble in one another  Water & oil

8. Dissolving Process   Surface Area = faster dissolving Loose sugar dissolves faster than sugar cube  Stir/Shake = faster dissolving Allows molecules to interact with one another  heat = faster dissolving  Solutes dissolve faster in warmer liquids   heat  speed  collisions and more transfer of energy

9. Unsaturated solution = soln that is able to dissolve more solute Saturated solution = soln that can not dissolve any more solute

10. Factors affecting solubility Temperature – as Temp  solubility   Super-saturated solution = soln holding more dissolved solute than is specified by its solubility at a given temperature Pressure – strongly influences solubility of gases. Solubility  as the partial pressure of the gas above the solution   Carbonated beverages are bottled under  pressure of CO 2 when opened pressure  above liquid and bubbles are released

11. 16.1 Concentrations of solutions A measure of the amount of solute that is dissolved in a given quantity of solvent  Dilute – one that contains a small amount of solute  Concentrated – one that contains a large amount of solute Molarity – the number of moles of solute dissolved in one liter of solution  M = Moles solute/liters of solution  Units = moles/L

13. Calculating molarity A saline solution contains 0.900g NaCl in exactly 100.0ml of solution what is the molarity of the solution? Known  0.90g NaCl (molar mass 58.453g/mol) 0.90g X (1mol/58.453g) = 0.0154 mol  100 mL = 0.1L Use equation M = moles/liter  0.0154mol/0.1L = 0.154M

14. Calculation 1 A solution has a volume of 250mL and contains 0.70mol NaCl. What is its molarity? Known  0.70mol NaCl  250mL = 0.250L Use equation M = moles/liter  M = 0.70mol/0.250L = 2.8M

15. Calculation 2 How many moles of ammonium nitrate are in 335mL of 0.425M NH 4 NO 3 ? Known  M = 0.425M  335mL = 0.335L Manipulate equation to solve for moles Moles = Molarity X Liters Moles = 0.425M X 0.335L = 0.142 moles

16. Dilutions Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change  Moles solute before = moles solute after  M 1 V 1 = M 2 V 2

17. Preparing a dilute solution How many milliliters of 2.00M MgSO 4 solution must be diluted with water to prepare 100.00mL of 0.400M MgSO 4 ? Known   M 1 = 2.00M  M 2 = 0.4M  V 2 = 100mL Manipulate equation to solve for V 1 V 1 = (M 2 V 2 )/M 1 V 1 = (0.40M X 100.0mL)/2.00M = 20.0mL

18. Calculation 1 How many milliliters of 4.00M KI are needed to prepare 250mL of 0.760M KI? Known  M 1 = 4.00M  M 2 = 0.760M  V 2 = 250mL Manipulate equation to solve for V 1 V 1 = (M 2 V 2 )/M 1 V 1 = (0.760M X 250mL) / 4.00M = 47.5 mL

19. Calculation 2 How many milliliters of 0.20M NaCl are formed when you dilute a 50mL solution of 1.0M NaCl using water? Known  V1 = 50mL  M2 = 0.20M  M1 = 1.0M Manipulate equation to solve for V2 V 2 = (M 1 V 1 )/M 2 V 2 = (1.0M X 50mL) / 0.20M = 250mL

20. Percent solutions The concentration of a solution in percent can be expressed in two ways  Ratio of volume of solute to the volume of the solution (% (v/v) % volume = volume solute/volume soln X 100%  Ratio of the mass of the volume to the mass of the solution (% (m/m) % mass = mass of solute/mass soln X 100%

21. Calculating percent solution (v/v) What is the percent by volume of ethanol (C 2 H 6 O) in the final solution when 85mL of ethanol is diluted to a volume of 250mL with water? Known  Volume ethanol 85mL  Volume of solution 250mL % v/v = Volume solute/volume solution X 100% % v/v = (85mL/250mL) X 100% = 34%

22. Calculation 1 If 10mL of acetone (C 3 H 6 O) is diluted with water to a total solution volume of 200mL, what is the percent by volume of the solution? Known  Volume of acetone 10mL  Volume of solution 200mL % v/v = Volume solute/volume solution X 100% % v/v = (10mL/200mL) X 100% = 5%

23. Calculation 2 (m/m) How many grams of K 2 SO 4 would you need to prepare 1500g of 5% K 2 SO 4 ? Known  Mass of solution 1500g  % by mass 5% Manipulate equation to solve for mass solute mass solute = mass solution X % by mass Mass solute = 1500g X 5% = 75g