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Homogeneous mixture = solution Solute the substance that’s being dissolved Solvent - the substance that the solute is dissolved in.

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Presentation on theme: "Homogeneous mixture = solution Solute the substance that’s being dissolved Solvent - the substance that the solute is dissolved in."— Presentation transcript:

1 Homogeneous mixture = solution Solute the substance that’s being dissolved Solvent - the substance that the solute is dissolved in.

2 Colligative properties: properties of solutions that depend only on the NUMBER of solute particles present. The chemical nature of the solute particles does not affect the colligative properties. == Solution properties that are affected: freezing point, boiling point and osmotic pressure of the SOLVENT.

3 Particles in solution If an ionic compound is used as the solute, it will break up into its ions. Notice how the subscripts in the formula for the ionic solids become coefficients for the ions. Total #particles NaCl (s)Cl -1 (aq)Na +1 (aq) + 2 MgCl 2 (s)Cl -1 (aq)Mg +2 (aq) + 2 3 AlCl 3 (s)Al +3 (aq) + Cl -1 (aq) 3 4 C 6 H 12 O 6 (s) Remember that molecular (covalent) compounds exist as discrete particles…they don’t break up when they dissolve. C 6 H 12 O 6 (aq) 1

4 Colligative properties Freezing point depression: The normal freezing point of the solvent decreases as the number of solute particles increases. Example: antifreeze Freezing point  T f = K f m  T f = the change in the freezing point of the solvent K f = molal freezing point constant (depends on solvent) m = MOLALITY of the solution = Moles of solute particles Kg of solvent

5 Boiling point elevation: The normal boiling point of the solvent increases as the number of solute particles increases. Example: engine coolant Boiling point  T b = K b m  T b = the change in the boiling point of the solvent K b = molal boiling point constant (depends on solvent) Ex: What will be the boiling point of a 0.1 m aqueous solution of a nonelectrolyte solute? (K b of H 2 O = 1.86  C/m)  T b = K b m  T b = (1.86  C/m)(0.1 m)  T b = 0.186  C New boiling point =100  C + 0.186  C = 100.186  C

6 Vapor Pressure Lowering: Vapor pressure in a closed container is caused by liquid particles that have escaped from the liquid and are now in a gas state. Adding a nonvolatile solute to the liquid causes a DECREASE in the vapor pressure of the solvent. Since the extent of vapor pressure depends only on the number of solute particles that have been added, vapor pressure lowering is another colligative property.

7 Which solute would cause the greatest freezing point depression? A) 1 g of NaClB) 2 g of NaCl C) 3 g of NaClD) 4 g of NaCl A) 1 mol NaClB) 1 mol MgCl 2 C) 1 mol KClD) 1 mol AlCl 3 A) 1 mol of a nonelectrolyte*B) 0.4 mol MgCl 2 C) 0.25 mol KClD) 0.4 mol NaCl *a nonelectrolyte is one that doesn’t break apart in solution.

8 molality = Ex: What is the molality of a solution prepared by dissolving 1.5 mol H 2 O in 500 mL of ethanol (density = 0.789 g/mL)? m = kg n Step 1: Find the mass of ethanol mass = DV = (0.789 g/mL)(500 mL) = 394.5 g of ethanol = 0.3945 kg ethanol Step 2: Calculate the molality


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