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CONCEPT: PHASE DIAGRAMS Under appropriate conditions of pressure and temperature, most substances can exist in 3 states of matter:, and. Microscopic Explanation for the Behavior of Gases, Liquids and Solids Gas Liquid Solid Assumes the and of its container. Assumes the of the portion of its container it occupies, but not the. Maintains a fixed and compressible compressible compressible Viscosity Viscosity Viscosity Viscous Viscous Viscous Now, a convenient way to show the effect that temperature and pressure has on a pure substance in a closed system without any air is to use a phase diagram. Page 2
PRACTICE: PHASE DIAGRAMS a) At what temperature can we no longer tell the difference between the liquid and the gas? b) Which point represents an equilibrium between the solid, liquid and gas phase? c) Which line segment represents an equilibrium between fusion and freezing? d) Which line segment represents an equilibrium between sublimation and deposition? e) Which line segment represents an equilibrium between condensation and vaporization? f) What is the normal freezing point of this unknown substance? g) What is the normal boiling point of this unknown substance? Page 3
CONCEPT: HEATING & COOLING CURVES In heating and cooling curves we have the representation of the amount of heat absorbed or released during phase changes. Heating Curve Temperature ( o C) Time Cooling Curve Specific Heat of Ice 2.09 ΔH Fusion 334 J g Specific Heat of Water 4.184 ΔH Vaporization 2260 J g J g o C J g o C Specific Heat of Steam 1.84 J g o C Temperature ( o C) Time Page 4
PRACTICE: HEATING & COOLING CURVE CALCULATIONS 1 EXAMPLE: How much energy (kj) is required to convert a 76.4 g acetone (molar mass = 58.08 g/mol) as a liquid at -30 o C to a solid at -115.0 o C? J Specific Heat of Solid 1.65 g o C ΔH Fusion 7.27 kj mol J Specific Heat of Liquid 2.16 g o C J Specific Heat of Gas 1.29 g o C T Melting 95.0 o C Page 5
PRACTICE: HEATING & COOLING CURVE CALCULATIONS 2 PRACTICE: If 53.2kJ of heat are added to a 15.5g ice cube at - 5.00 o C, what will be the resulting state and temperature of the substance? Specific Heat of Ice 2.09 J g o C ΔH Fusion 334 J g Specific Heat of Water 4.184 J g o C ΔH Vaporization 2260 J g Specific Heat of Steam 1.84 J g o C Page 6
CONCEPT: CLASIUS-CLAPEYRON EQUATION By using the Clasius-Clapeyron equation a quantitative relationship between and can be established. ln P 2 = ΔH vap P 1 R # 1 1 & % ( $ T 2 T 1 ' ` EXAMPLE : The heat of vaporization ( Hvap) of water is 40.3 kj/mol at its normal boiling point at 100 o C. What is the vapor pressure (in mmhg) of water at 60 o C? Page 7
CONCEPT: MOLARITY Molarity (M) can serve as the connection between the interconversion of to and vice versa. For example, a 5.8 M NaCl solution really means per. ( Molarity = MolesSolute ) (LitersSolution) A typical mixture consists of a smaller amount of one substance, the, dissolved in a larger amount of another substance, the. Together they form a. Page 8
PRACTICE: MOLARITY EXAMPLE 1: 2.64 grams of an unknown compound was dissolved in water to yield 150 ml of solution. The concentration of the solution was 0.075 M. What was the molecular weight of the substance? ( M = MolesSolute ) (Liters Solution) EXAMPLE 2: A solution is prepared by dissolving 0.1408 mol calcium nitrate, Ca(NO3)2, in enough water to make 100.0 ml of stock solution. If 20.0 ml of this solution is then mix with an additional 90 ml of deionized water, calculate the concentration of the calcium nitrate solution. PRACTICE 1: What is the molarity of calcium ions of a 650 ml solution containing 42.7 g of calcium phosphate? PRACTICE 2: A solution with a final volume of 750.0 ml was prepared by dissolving 30.00 ml of benzene (C6H6, density = 0.8787 g ) in dichloromethane. Calculate the molarity of benzene in the solution. ml Page 9
CONCEPT: SOLUTIONS AND INTERMOLECULAR FORCES Molarity (M) can serve as the connection between the interconversion of to and vice versa. For example, a 5.8 M NaCl solution really means per. Molarity = Moles of Solute Liters of Solution A typical mixture consists of a smaller amount of one substance, the, dissolved in a larger amount of another substance, the. Together they form a. According to the theory of likes dissolves likes compounds with the same or will dissolve into each other. EXAMPLE: Butane, a nonpolar organic compound, is most likely to dissolve in a. HCl b. C6H5OH c. C8H18 d. AlCl3 e. What the heck is butane? Page 10
CONCEPT: CALCULATE MOLARITY Molarity is the concentration of a solution represented as moles of solute per liter of solution: Molarity (M) = Moles of solute Liters of solution Understanding this explanation for molarity allows you to further expand it. For example, if we are given 0.20 M NaOH this means: Osmolarity 0.20 M NaOH= 0.20 mole NaOH 1Liter of solution Ionic molarity or osmolarity represents the molarity of dissolved ions in a solution. Dissolution of Ionic Compound 0.10 M KNO 3 K + + NO 3 Osmolarity Osmolarity = ( # of ions) x (M of Compound) 2 Ions Dilutions Another common idea related to molarity deals with the dilution of stock solutions. In a dilution a concentrated solution is made more diluted by adding water. M 1 V 1 = M 2 V 2 M 1 = Molarity before diltution V 1 = Volume before dilution M 2 = Molarity after dilution M 1 is larger than M 2 V 2 is larger than V 1 V 2 = V 1 + V Water Added V 2 = Volume after dilution Page 11
PRACTICE: CALCULATE MOLARITY CALCULATIONS 1 EXAMPLE 1: Stock hydrochloric acid solution is 21.0% by mass HCl and has a specific gravity of 1.75. What is the molarity of the solution? EXAMPLE 2: What is the concentration of nitrate ions in a solution that contains 83.3 g lead (IV) nitrate, Pb(NO3)4, dissolved in 700 ml solution? MW of Pb(NO3)4 is 455.24 g/mol. EXAMPLE 3: How many micrograms of K2CO3 are required to prepare 120 ml of 0.325 M K2CO3? MW of K2CO3 is 138.21 g/mol. Page 12
PRACTICE: CALCULATE MOLARITY CALCULATIONS 2 EXAMPLE 1: If 50.0 ml of water is added to 80.0 ml of a 6.00 M solution, what will be the new concentration of the solution? EXAMPLE 2: To what volume should you dilute 72.93 ml of a 7.505 M LiCl solution so that 23.25 ml contains 2.25 g LiCl? MW of LiCl is 42.392 g/mol. EXAMPLE 3: 870.0 g of water contains 0.990 g of sodium phosphate, Na3PO4. Determine the concentration of Na3PO4 if the density of the solution is 1.10 g/ml. MW of Na3PO4 is 163.94 g/mol. Page 13
CONCEPT: MOLALITY Molality is depicted as moles of solute per kilograms of solvent: Molality (m) = Moles of solute kg of solvent In the same way we can expand molarity the same approach can be applied to molality: Osmolality 0.30 mnacl = 0.30 mole NaCl 1kg of solvent Ionic molality or osmolality represents the molality of dissolved ions in a solution. Dissolution of Ionic Compound 0.30 m NaCl Na + + Cl Osmolality Osmolality = ( # of ions) x (m of Compound) 2 Ions EXAMPLE: A solution is prepared by dissolving 43.0 g potassium chlorate, KClO3, in enough water to make 100.0 ml of solution. If the density of the solution is 1.760 g/ml, what is the molality of KClO3 in the solution? MW of KClO3 is 122.55 g/mol. Page 14
PRACTICE: MOLALITY CALCULATIONS EXAMPLE 1: If the molality of glucose, C6H12O6, in an aqueous solution is 2.56 what is the molarity? Density of the solution is 1.530 g/ml. EXAMPLE 2: What is the ionic molality of nitrate ions in 0.305 m lead (IV) nitrate, Pb(NO3)4?. PRACTICE: What is the mass percent of NH3 of a 1.25 m aqueous solution of NH3? Page 15
CONCEPT: MOLE FRACTION Mole Fraction is depicted as moles of solute per moles of solution: Mole Fraction(X) = Mole of Solute Mole of Solution = For example, if we are given a mole fraction of 0.160 for NaOH this means: 0.160 = 0.160 mole NaOH 1.0 mole Solution = EXAMPLE 1: If the mole fraction of glucose, C6H12O6, in an aqueous solution is 0.320 what is the molarity? Density of the solution is 1.530 g/ml. EXAMPLE 2: If the mole fraction of methanol, CH3OH, in an aqueous solution is 0.060 what is the molality? Density of the solution is 1.39 g/ml. EXAMPLE 3: Calculate the mole fraction of acetic acid, HC2H3O2, in a 27.13 mass % aqueous solution (d = 0.9883 g/ml). MW of HC2H3O2 is 60.054 g/mol Page 16
CONCEPT: MASS PERCENT Mass or weight percent is the percentage of a given element or compound within a solution. Mass Percent = For example, if we are given 23.0% NaOH this means: 23.0 %NaOH= 23.0 grams NaOH 100 grams Solution Mass Component Total Mass 100 EXAMPLE 1: Calculate the amount of water (in kilograms) that must be added to 12.0 g of urea, (NH2)2CO, in the preparation of a 18.3 percent by mass solution. The molar mass of urea, (NH2)2CO, is 60.055 g/mol. EXAMPLE 2: A solution was prepared by dissolving 51.0 g of KBr in 310 ml of water. Calculate the mass percent of KBr in the solution. PRACTICE: At 298 K, a solution is prepared by dissolving 12.7 g NaCl in 95.5 ml of water. What is the ppm of NaCl if the density of water at this temperature is 0.9983 g/ml. Page 17
CONCEPT: PROPERTIES OF SOLUTIONS The 4 properties help to explain what happens to a pure solvent as we add solute to it. EXAMPLE: Explain what happens to each of the following properties as solute is added to a pure solvent. a. Boiling Point ΔT b = i k b m b. Freezing Point ΔT f = i k f m c. Osmotic Pressure Π = i MRT d. Vapor Pressure o P Solution = X Solvent P Solvent EXAMPLE: Which of the following compounds will have the highest boiling point? a) 0.10 m sucrose b) 0.10 m CsBrO3 c) 0.35 m CH3OH d) 0.15 m SrBr2 Page 18
PRACTICE: PROPERTIES OF SOLUTIONS EXAMPLE 1: Pure water boils at 100 o C. What is the expected boiling point of water after the addition of 13.12 g calcium bromide, CaBr2, to 325 g water. Kb = 0.512 o C/m. (MW of CaBr2 is 199.88 g/mol). EXAMPLE 2: The vapor pressure of water at 100.0 o C is 0.630 atm. Determine the amount (in grams) of aluminum fluoride, AlF3, (in grams) needed to reduce its vapor pressure to 0.550 atm. (MW of AlF3 is 83.98 g/mol). PRACTICE 1: Beta-carotene is the most important of the A vitamins. Calculate the molar mass of Beta-carotene if 25.0 ml of a solution containing 9.88 mg of Beta-carotene has an osmotic pressure of 56.16 mmhg at 30 o C. Page 19
CONCEPT: THE LIQUID STATE VAPOR PRESSURE is defined as the partial pressure of vapor molecules above the surface of the liquid under the dynamic equilibrium condition of condensation and evaporation. EXAMPLE: The vapor pressure of pure liquid A is 550 torr and the vapor pressure of pure liquid B is 320 torr at room temperature. If the vapor pressure of a solution containing A and B is 465 torr, what is the mole fraction of A in the solution? Page 20
CONCEPT: THE LIQUID STATE (CALCULATIONS) EXAMPLE: Determine the vapor pressure lowering associated with 1.32 m C6H12O6 solution (MW: 180.156 g/mol) at 25 o C. PRACTICE: The following boiling points belong to one of the following compounds: 117 o C, 78 o C, 34.5 o C & 23 o C CH3-O-CH3 CH3CH2OH CH3CH2-O-CH2CH3 CH3CH2CH2CH2OH a) Which boiling point goes with what compound? b) If each of the following substances were placed in separate sealed clear bottles at room temperature, could you identify one of the substances right away? Page 21
CONCEPT: SOLUBILITY AS AN EQUILIBRIUM PROCESS When an ionic solid dissolves, ions leave the solid and become dispersed in the solvent. In a(n) solution the maximum amount of dissolved solute is present in the solvent. In a(n) solution additional amounts of solute can be further dissolved in the solvent. In a(n) solution more than the equilibrium concentration of solute has been dissolved. EXAMPLE C1: Caffeine is about 10 times as soluble in warm water as in cold water. A student puts a hot-water extract caffeine into an ice bath, and some caffeine crystallizes. What is the identity of the solution before it s been placed in an ice bath? a) Saturated b) Super saturated c) Unsaturated d) Not enough information to answer the question. law explains the relationship between gas pressure and solubility: the solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution: S Gas = k H P Gas Page 22
PRACTICE: SOLUBILITY AS AN EQUILIBRIUM PROCESS EXAMPLE: Henry s Law Constant for nitrogen in water is 1.67 x 10-4 M atm -1. If a closed canister contains 113 ppb nitrogen, what would be its pressure in atm? PRACTICE 1: In general, as the temperature increases, the solubility of gas in a given liquid, and the solubility of most solids in a given liquid. a. Increases, increases b. Increases, decreases c. Decreases, increases d. Decreaes, decreases PRACTICE 2: At a partial pressure of acetylene 1.35 atm, 1.21 moles of it dissolves in 1.05 L of acetonitrile. If the partial pressure of acetylene in acetone is increased to 12.0 atm, then what is its solubility? Page 23