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Physical Properties of Solutions Unit 10 Why are some compounds more effective in melting ice than others?

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Presentation on theme: "Physical Properties of Solutions Unit 10 Why are some compounds more effective in melting ice than others?"— Presentation transcript:

1 Physical Properties of Solutions Unit 10 Why are some compounds more effective in melting ice than others?

2 Solutions in the World Around Us

3 Definitions Review from Chapter 4: Solution: homogeneous mixture of two or more substances. Solute: the substance being dissolved. Solvent: the substance doing the dissolving

4 Colloid – solution of solid particles small enough to be suspended in solution Example: milk, paint, smoke Colloids may look clear or almost clear when dilute enough

5 Tyndall effect – scattering of light by particles in a colloid or suspension ▫Causes the beam of light to become visible ▫Why you can see rays from the sun (particles in the air scatter the light)

6 Suspension – a mixture from which some of the particles settle out slowly upon standing ▫Particles are too big to be dissolved ▫Example: sand in water, dust in air ▫Suspensions can be filtered

7 Concentration Concentration of a solution: the quantity of a solute in a given quantity of solution (or solvent). ▫A concentrated solution contains a relatively large amount of solute vs. the solvent ▫A dilute solution contains a relatively small concentration of solute vs. the solvent “Concentrated” and “dilute” aren’t very quantitative.

8 Need different concentration units to describe different properties of solutions: Molarity (review) Mass Percent Molality Solution Concentration

9 Example 1: What is the molarity of a solution made by dissolving 12.5 g of oxalic acid in 456 mL of solution? Molarity (M)

10 Example 2: How many grams of sodium carbonate are needed to prepare.250 L of an aqueous 0.300 M soln.? Molarity (M)

11 Mass Percent Ex. 4: What is the % by mass of the solute in a soln. made by adding 1.20 g of methyl alcohol to 16.8 g of H 2 O?

12 Molality (m) Example 5: A solution contains 15.5 g of urea, NH 2 CONH 2, in 74.3 g of water. Calculate the molality of the urea.

13 Why use Molality?? Molality (m) is the number of moles of solute per one kilogram of solvent (not solution!) EOS Molality does NOT change with temperature! Molarity (M) varies with temperature due to the expansion or contraction in the volume of the solution 

14 Dilution of Solutions Example 6: How would you prepare 0.250 L of 0.300 M Na 2 CO 3 starting with 1.33 M solution?

15 Vocabulary of Solutions Liquids that mix in all proportions are called miscible.

16 When there is a dynamic equilibrium between an undissolved solute and a solution, the solution is saturated EOS A saturated solution is one in which a given amount of solvent has dissolved the absolute maximum amount of solute at that temperature.

17 Vocabulary EOS A solution which contains less solute than can be held at equilibrium is unsaturated

18 Supersaturation A supersaturated solution is created when a warm, saturated solution is allowed to cool without the precipitation of the excess solute Testing for saturation: add crystal of a solid and watch for crystallization.

19 Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.” Sodium acetate has an ENDOthermic heat of solution.

20 Supersaturated Sodium Acetate Sodium acetate has an ENDOthermic heat of solution. NaCH 3 CO 2 (s) + heat ----> Na + (aq) + CH 3 CO 2 - (aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na + (aq) + CH 3 CO 2 - (aq) ---> NaCH 3 CO 2 (s) + heat

21 Factors Which Influence Solubility Temperature Pressure (for gases) Surface Area (Impacts how QUICKLY a substance will dissolve)

22 Solubility Graphs or Curves The concentration of the solute in a saturated solution is the solubility of the solute, and may be shown on a graph or curve.

23 Solubility Curves Example: What mass of solute will dissolve in 100mL of water at the following temperatures. 1.KNO 3 at 70°C 2.NaCl at 100°C

24 Solubility Curves Example: What term - saturated, unsaturated, or supersaturated – best describes: A solution that contains 70g of NaNO 3 per 50 g H 2 O at 30°C

25 Solubility Curves Example: What term - saturated, unsaturated, or supersaturated – best describes: A solution that contains 70 g of dissolved KCl per 200 g H 2 O at 80°C.

26 Solubility Curves Example: Determine the molality of a saturated NaCl solution at 25°C.

27 Ionic Compounds EOS The attractions of water dipoles for ions pulls the ions out of the crystalline lattice and into aqueous solution.

28 The Solubilities of Gases Most gases become less soluble in liquids as the temperature increases Example: can of soda going flat on a hot day

29 Effect of Pressure When P gas drops, solubility drops. Also known as Henry’s Law.

30 Colligative Properties When adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreases Melting point decreases Boiling point increases Osmosis is possible (osmotic pressure changes) These changes are called COLLIGATIVE PROPERTIES.

31 Colligative Properties Definition: Colligative Properties depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND or IDENTITY of solute particles.

32 FP Depression and BP Elevation— Two Colligative Properties The presence of the solute lowers (depresses) the freezing point of the solvent (  T f ) and increases (elevates) the boiling point of the solvent (  T b ) EOS Examples: adding salt to water allows the water temperature to exceed 100 o C, thereby cooking food faster; placing salt on ice lowers the freezing point, allowing ice to melt at a lower temperature.

33 Elevation of Boiling Point Elevation in BP = ∆T b = K b m i k b = constant; characteristic of solvent i = van’t Hoff factor = 1 FOR ALL NONELECTROLYTES! m = molality ∆T b = number of degrees BP goes UP

34 Lowering (depression) of Freezing Point Depression in FP = ∆T f = K f m i k f = constant; characteristic of solvent i = van’t Hoff factor = 1 FOR ALL NONELECTROLYTES! m = molality ∆T f = number of degrees FP goes DOWN

35 Sample Problem Example: Calculate the boiling point of solution that contains 50.0 g of glucose, C 6 H 12 O 6, in 400 g of water. The molal boiling point constant of water is 0.52 o C/m. ▫Calculate the molality first! ▫i = 1 since glucose is a NONELECTROLYTE.

36 Freezing Point Depression Freezing point of a solution is lower than the freezing point of the pure solvent. Dissolving substances lowers the freezing point of a solvent. Ex: Icy pavement - throw down CaCl 2 or NaCl, and the water will then freeze at a lower temperature

37 Ex: Antifreeze: a solution of ethylene glycol in water 1. Prevents car’s radiator from freezing in the winter. 2. Prevents car’s radiator from boiling over in the summer The more ethylene glycol in the water, the lower the freezing point, and the higher the boiling point.

38 Freezing Point Depression The freezing point of a solution is LOWER than that of the pure solvent. FP depression = ∆T FP = K FP m Pure water Ethylene glycol/water solution

39 Colligative Properties of Electrolytes Nonelectrolytes vs. electrolytes Nonelectrolytes produce only molecules in solution; electrolytes produce ions. NaCl  Na + + Cl – The greater the product of molality and number of ions, the greater the boiling point elevation or freezing point depression.

40 Modified Equation: Van ’ t Hoff Factor ∆T = K m i i = van’t Hoff factor = number of particles produced per formula unit. Compound Theoretical Value of i Sugar (NONELECTROLYTE)1 NaCl2 CaCl 2 3

41 Sample Problem Example. Rank the following aqueous solutions in order of lowest to highest melting point: ▫0.010 m C 6 H 12 O 6 ▫0.0050 m MgCl 2 ▫0.0080 m HCl ▫0.0040 m Al 2 (SO 4 ) 3


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