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Chapter 15: Solutions. Solutions Solution – a homogeneous mixture of two or more substances in a single physical state. Solution – a homogeneous mixture.

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Presentation on theme: "Chapter 15: Solutions. Solutions Solution – a homogeneous mixture of two or more substances in a single physical state. Solution – a homogeneous mixture."— Presentation transcript:

1 Chapter 15: Solutions

2 Solutions Solution – a homogeneous mixture of two or more substances in a single physical state. Solution – a homogeneous mixture of two or more substances in a single physical state. Properties of solutions – Properties of solutions – –The particles are very small (atoms, molecules or ions) –The particles in a solution are evenly distributed or uniformly mixed (a spoonful of lemonade tastes the same as the whole glass)

3 Parts of solutions Solute – the thing that gets dissolved Solute – the thing that gets dissolved Solvent – the thing that does the dissolving Solvent – the thing that does the dissolving Soluble – something that can be dissolved (salt and sugar are soluble in water) Soluble – something that can be dissolved (salt and sugar are soluble in water) Insoluble – a substance that cannot be dissolved (Hg and oil) Insoluble – a substance that cannot be dissolved (Hg and oil)

4 Types of solutions Solid Solutions – alloys – made by melting different metals and then cooling them Solid Solutions – alloys – made by melting different metals and then cooling them Makes material stronger, higher melting points and greater resistance to corrosion Makes material stronger, higher melting points and greater resistance to corrosion Ex. Dental filling – mercury in silver Ex. Dental filling – mercury in silver Sterling silver – copper in silver 18 karat gold – gold, copper and silver Coinage bronze – copper, tin and zinc

5 Types of solutions Gaseous solutions – air we breathe. Nitrogen with oxygen in it Gaseous solutions – air we breathe. Nitrogen with oxygen in it If gas molecules mix they become a solution. Anytime gases are near each other they will mix, therefore any gas mixture is a solution. If gas molecules mix they become a solution. Anytime gases are near each other they will mix, therefore any gas mixture is a solution.

6 Types of solutions Liquid Solutions – MOST COMMON! the solvent and the solution are both liquids. The solute can be a gas, liquid or solid. Liquid Solutions – MOST COMMON! the solvent and the solution are both liquids. The solute can be a gas, liquid or solid. Miscible – liquids that can be mixed in any amount (water and ethanol) Miscible – liquids that can be mixed in any amount (water and ethanol) Immiscible – liquids that cannot mix in any proportion (oil and water) Immiscible – liquids that cannot mix in any proportion (oil and water)

7 Types of solutions Aqueous solutions – solutions with water as the solvent Aqueous solutions – solutions with water as the solvent 2 types… 2 types… 1. When ionic compounds dissolve, ions are present and make an Electrolyte. (NaCl) 2. When molecular compounds dissolve, no ions are present and make a nonelectrolyte. (Sugar)

8 15-2 Concentration of Solutions Concentration – the amount of solute in a given amount of solvent. Concentration – the amount of solute in a given amount of solvent.

9 Molarity (M) Molarity = moles of solute Molarity = moles of solute Liters of solution Liters of solution Example problem… Example problem… What is the Molarity of a NaOH solution if 10.0g of NaOH is dissolved in enough solvent to make 0.100L of solution?

10 You try it!!! Find the Molarity of a solution formed by mixing 10.0g of H 2 SO 4 with enough water to make 100.0mL of solution Find the Molarity of a solution formed by mixing 10.0g of H 2 SO 4 with enough water to make 100.0mL of solution 10.0g H 2 SO 4 x 1 mol H 2 SO 4 = 0.102 mol 98.1 g H 2 SO 4 100.0 mL x 1 L = 0.100 L 1000 mL M = moles = 0.102 mol = 1.02 mol/L or 1.02 M liters 0.100 L

11 Molality (m) Molality = moles of solute Molality = moles of solute kg of Solvent kg of Solvent If 18.0g of C 6 H 12 O 6 is dissolved in 1kg of water. If 18.0g of C 6 H 12 O 6 is dissolved in 1kg of water. 18.0g C 6 H 12 O 6 x 1mol C 6 ­H 12 O 6 = 0.100 mol C 6 ­H 12 O 6 180g C 6­ H 12 O 6 m = 0.100 mol = 0.100 mol/kg or 0.100 m 1 kg

12 Mole Fraction ( χ ) Mole Fraction = moles of component Mole Fraction = moles of component Total moles of solution Total moles of solution What is the mole fraction of SO 2 in a gas containing 128.0g of SO 2 dissolved in every 1500.g of CO 2 What is the mole fraction of SO 2 in a gas containing 128.0g of SO 2 dissolved in every 1500.g of CO 2

13 change grams to moles change grams to moles 128g SO 2 x 1 mole SO 2 = 2.0 moles SO 2 128g SO 2 x 1 mole SO 2 = 2.0 moles SO 2 64.0 g SO 2 64.0 g SO 2 1500 g CO 2 x 1 mole CO 2 = 34.1 moles CO 2 1500 g CO 2 x 1 mole CO 2 = 34.1 moles CO 2 44.0 g CO 2 44.0 g CO 2 χ SO 2 = moles of SO 2 χ SO 2 = moles of SO 2 Total moles of solution Total moles of solution χ SO 2 = 2.0moles SO 2 (2.0 moles SO 2 + 34.1 moles CO 2 ) (2.0 moles SO 2 + 34.1 moles CO 2 ) χ SO 2 = 0.0554 χ SO 2 = 0.0554

14 You try it!!! A gas mixture contains 50.4 g of N 2 O and 65.2g of O 2 what is the mole fraction of N 2 O? A gas mixture contains 50.4 g of N 2 O and 65.2g of O 2 what is the mole fraction of N 2 O? (0.360) (0.360)

15 15-3 The Formation of Solutions Why does salt dissolve? Why does salt dissolve? –The particles are attracted to the water (opposite dipoles attract to ions). –The water attaches itself on the face of NaCl and pulls it apart. –Then the water surrounds the ions. Solvation - the interaction between solute and solvent Solvation - the interaction between solute and solvent Hydration – when the solvent is water Hydration – when the solvent is water

16 Solvation and energy Energy is needed when bonds are broken between the solute and solvent Energy is needed when bonds are broken between the solute and solvent When the solvent and solute are attracted energy is released. When the solvent and solute are attracted energy is released. Net result is either endothermic or exothermic Net result is either endothermic or exothermic –Cold packs ex. NH 4 OH –Hot packs ex. Na 2 S 2 O 3

17 Solubility Solubility – the amount of solute that will dissolve in a specific solvent under given conditions. Solubility – the amount of solute that will dissolve in a specific solvent under given conditions. Depends on… Depends on… –Nature of solute and solvent – polar compounds dissolve polar compounds …“likes dissolve likes” –Temperature – As temperature increases, gas particles get more energy and solubility of a gas decreases (better to keep soda cold!)

18 Solubility (continued) –Solubility of a solid – as temp increases, solubility increases (sugar to iced tea or hot tea) Pressure – solubility of a gas is increases as the pressure increases because when the pressure is increased more gas can be dissolved Pressure – solubility of a gas is increases as the pressure increases because when the pressure is increased more gas can be dissolved –Scuba divers – nitrogen is dissolved in your blood because of the extra pressure, if they come up too fast the nitrogen will bubble out of your blood. (bends)

19 Saturated – if a solution contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure Saturated – if a solution contains as much solute as can possibly be dissolved under the existing conditions of temperature and pressure Unsaturated – a solution that has less than the maximum amount of solute Unsaturated – a solution that has less than the maximum amount of solute Saturation

20 Supersaturated – contains a greater amount of solute than needed to form a saturated solution (rock candy) Supersaturated – contains a greater amount of solute than needed to form a saturated solution (rock candy)

21 Solubility Curves Compares solubility to temperature Compares solubility to temperature Line shows a saturated amount Line shows a saturated amount Below line …unsaturated Below line …unsaturated Above line …supersaturated Above line …supersaturated

22 15-4 Colligative Properties Colligative Properties – depend on the collective effort of the solute properties and not their identity. Colligative Properties – depend on the collective effort of the solute properties and not their identity. –Ex. Vapor pressure reduces Vapor pressure reduces Boiling point elevates Boiling point elevates Freezing point reduces Freezing point reduces Osmotic pressure decreases Osmotic pressure decreases

23 Vapor pressure reduction When a nonvolatile solute is added to a solvent, the solute takes up space at the surface which prevents some of the solvent from evaporating. When a nonvolatile solute is added to a solvent, the solute takes up space at the surface which prevents some of the solvent from evaporating. Gases are still returning to the liquid at the same rate. This reduces the vapor pressure of the solution. Gases are still returning to the liquid at the same rate. This reduces the vapor pressure of the solution.

24 Antifreeze is added to a car to keep the water in the radiator from boiling. Antifreeze is a nonvolatile substance so it reduces the vapor pressure Antifreeze is added to a car to keep the water in the radiator from boiling. Antifreeze is a nonvolatile substance so it reduces the vapor pressure The boiling point increases because it takes more energy to reach atmospheric pressure The boiling point increases because it takes more energy to reach atmospheric pressure ∆T b = K b m ∆T b = K b m –∆T b = boiling point elevation –K b = molal boiling point constant (varies with solvent) –m= molality Boiling point elevation

25 Water with sugar added to it will boil at a higher temperature than pure water. By how much will the boiling point of water be elevated if 100g of C 12 H 22 O 11 is added to 500 g of water. K b = 0.52 C/m Water with sugar added to it will boil at a higher temperature than pure water. By how much will the boiling point of water be elevated if 100g of C 12 H 22 O 11 is added to 500 g of water. K b = 0.52 C/m 100.0 g sucrose x 1 mole sucrose =0.292 mol 342.3g sucrose m = 0.292 mol = 0.584 m 0.500kg H 2 O 0.500kg H 2 O ∆T b = K b m = 0.52 ⁰ C/m x 0.584 m = 0.30 ⁰ C ∆T b = K b m = 0.52 ⁰ C/m x 0.584 m = 0.30 ⁰ C Example Problem…

26 Freezing Point Depression the temp at which a liquid becomes a solid is decreased when there are solute particles because they get in the way of the attractive forces so the particles have to slow down more to freeze the temp at which a liquid becomes a solid is decreased when there are solute particles because they get in the way of the attractive forces so the particles have to slow down more to freeze Eq. ∆T f = K f m Eq. ∆T f = K f m –∆T f = freezing point depression –K f = molal freezing point constant (varies with solvent) –m= molality

27 Example Problem… Calculate the freezing point depression of a solution of 100.g of C 12 H 6 O 2 in 0.500Kg of water. The K f is 1.86C/m Calculate the freezing point depression of a solution of 100.g of C 12 H 6 O 2 in 0.500Kg of water. The K f is 1.86C/m Molality = mol solute/kg solution Molality = mol solute/kg solution 100.0 g C 2 H 6 O 2 x 1 mol = 1.61 mol 62.0g 100.0 g C 2 H 6 O 2 x 1 mol = 1.61 mol 62.0g m = 1.61 mol = 3.23 m 0.500Kg ∆T f = K f m = 1.86 ⁰ C/m x 3.23m = 6.01 ⁰ C


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