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Common Types of Solution

Common Types of Solution
Chapter 13 Solutions www.ThesisScientist.comSolutions • What is in a solution • How do we make them • How do we dilute them • Does pressure and temperature make a difference • How do solutions mix in our body www.ThesisScientist.comTragedy in Cameroon • Lake Nyos  Lake in Cameroon, West Africa.  On August 22, 1986, 1,700 people and 3,000 cattle died. • Released carbon dioxide cloud.  CO seeps in from underground 2 and dissolves in lake water to levels above normal saturation.  Though not toxic, CO is heavier 2 than air—the people died from asphyxiation. www.ThesisScientist.comTragedy in Cameroon: A Possible Solution • Scientists have studied Lake Nyos and similar lakes in the region to try and keep such a tragedy from reoccurring. • Currently, they are trying to keep the CO levels in the lake 2 water from reaching the very high supersaturation levels by venting CO from the lake 2 bottom with pipes. www.ThesisScientist.comSolutions • Homogeneous mixtures. Appears to be one substance, though really contains multiple materials. E.g., air and lake water. • Heterogeneous mixtures Appears to have two or more substances, you can see the individual components Chocolate chip cookies, people in a room, your neighborhood with houses, cars and trees www.ThesisScientist.comSolutions, Continued • Solute is the dissolved substance. Seems to “disappear.” “Takes on the state” of the solvent. • Solvent is the substance solute dissolves in. Does not appear to change state. • When both solute and solvent have the same state, the solvent is the component present in the highest percentage. • Solutions in which the solvent is water are called aqueous solutions. www.ThesisScientist.comCommon Types of Solution Solute Solvent Solution phase phase phase Example Air (mostly N and O ) Gaseous solutions Gas Gas 2 2 Gas Liquid Soda (CO in H O) 2 2 Vodka (C H OH in H O) Liquid solutions Liquid Liquid 2 5 2 Seawater (NaCl in H O) 2 Solid Liquid Brass (Zn in Cu) Solid solutions Solid Solid www.ThesisScientist.comSolubility • When one substance (solute) dissolves in another (solvent) it is said to be soluble. Salt is soluble in water. Bromine is soluble in methylene chloride. • When one substance does not dissolve in another it is said to be insoluble. Oil is insoluble in water. • The solubility of one substance in another depends on two factors: nature’s tendency towards mixing and the types of intermolecular attractive forces. www.ThesisScientist.comWill It Dissolve • Chemist’s rule of thumb: Like dissolves like • A chemical will dissolve in a solvent if it has a similar structure to the solvent. • When the solvent and solute structures are similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other. www.ThesisScientist.comClassifying Solvents Structural Solvent Class feature Water, H O Polar OH 2 Ethyl alcohol, C H OH Polar OH 2 5 Acetone, C H O Polar C=O 3 6 Nonpolar Toluene, C H CC and CH 7 8 Nonpolar Hexane, C H CC and CH 6 14 Nonpolar Diethyl ether, C H O CC, CH, 4 10 and CO www.ThesisScientist.comWill It Dissolve in Water • Ions are attracted to polar solvents.  Many ionic compounds dissolve in water. Generally, if the ions total charges 4. • Polar molecules are attracted to polar solvents.  Table sugar, ethyl alcohol, and glucose all dissolve well in water. Have either multiple OH groups or little CH. • Nonpolar molecules are attracted to nonpolar solvents. bcarotene (C H ) is not water soluble; it dissolves in fatty 40 56 (nonpolar) tissues. • Many molecules have both polar and nonpolar structures—whether they will dissolve in water depends on the kind, number, and location of polar and nonpolar structural features in the molecule. www.ThesisScientist.comSalt Dissolving in Water www.ThesisScientist.comSolvated Ions When materials dissolve, the solvent molecules surround the solvent particles due to the solvent’s attractions for the solute. This process is called solvation. Solvated ions are effectively isolated from each other. www.ThesisScientist.comPractice—Decide if Each of the Following Will Be Significantly Soluble in Water. • • p po otassiu tassium m io iod did ide, e, KI KI soluble. • • o octan ctane, e, C C H H insoluble. 8 8 18 18 • • m meth ethan ano ol, l, CH CH OH OH soluble. 3 3 • • co cop pp per, er, Cu Cu insoluble. • • cety cetyl l alcoh alcoho ol, l, CH CH (CH (CH ) ) CH CH OH OH insoluble. 3 3 2 2 14 14 2 2 • • iron iron(I (III II) ) su sulf lfid ide, e, Fe Fe S S insoluble. 2 2 3 3 www.ThesisScientist.comSupersaturated Solution A supersaturated solution has more dissolved solute than the solvent can hold. When disturbed, all the solute above the saturation level comes out of solution. www.ThesisScientist.comAdding Solute to various Solutions Unsaturated Saturated Supersaturated www.ThesisScientist.comElectrolytes • Electrolytes are substances whose aqueous solution is a conductor of electricity. • In strong electrolytes, all the electrolyte molecules are dissociated into ions. • In nonelectrolytes, none of the molecules are dissociated into ions. • In weak electrolytes, a small percentage of the molecules are dissociated into ions. www.ThesisScientist.comSolubility and Temperature • The solubility of the solute in the solvent depends on the temperature.  Higher temperature = Higher solubility of solid in liquid.  Lower temperature = Higher solubility of gas in liquid. www.ThesisScientist.comChanging Temperature = Changing Solubility • When a solution is saturated, it is holding the maximum amount of solute it can at that temperature. • If the temperature is changed, the solubility of the solute changes. If a solution contains 71.3 g of NH Cl in 100 g 4 of water at 90 C, it will be saturated. If the temperature drops to 20 C, the saturation level of NH Cl drops to 37.2 g. 4 Therefore, 24.1 g of NH Cl will precipitate. 4 www.ThesisScientist.comPurifying Solids: Recrystallization • Formation of the crystal lattice tends to reject impurities. • To purify a solid, chemists often make a saturated solution of it at high temperature; when it cools, the precipitated solid will have much less impurity than before. www.ThesisScientist.comSolubility of Gases: Effect of Temperature • Many gases dissolve in water. However, most have very limited solubility. • The solubility of a gas in a liquid decreases as the temperature increases. Bubbles seen when tap water is heated (before the water boils) are gases that are dissolved, coming out of the solution. Opposite of solids. www.ThesisScientist.comSolubility of Gases: Effect of Pressure • The solubility of a gas is directly proportional to its partial pressure. Henry’s law. The solubility of solid is not effected by pressure. • The solubility of a gas in a liquid increases as the pressure increases. www.ThesisScientist.comSolubility and Pressure • The solubility of gases in water depends on the pressure of the gas. • Higher pressure = higher solubility. www.ThesisScientist.comSolubility and Pressure, Continued When soda pop is sealed, the CO is under pressure. 2 Opening the container lowers the pressure, which decreases the solubility of CO and causes bubbles to form. 2 www.ThesisScientist.comSolution Concentrations www.ThesisScientist.comDescribing Solutions • Solutions have variable composition. • To describe a solution, you need to describe both the components and their relative amounts. • Dilute solutions have low amounts of solute per amount of solution. • Concentrated solutions have high amounts of solute per amount of solution. www.ThesisScientist.comConcentrations—Quantitative Descriptions of Solutions • A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution. • Concentration = Amount of solute in a given amount of solution. Occasionally amount of solvent. www.ThesisScientist.comMass Percent • Parts of solute in every 100 parts solution.  If a solution is 0.9 by mass, then there are 0.9 grams of solute in every 100 grams of solution. Or 10 kg solute in every 100 kg solution. • Since masses are additive, the mass of the solution is the sum of the masses of solute and solvent. Mass of Solute, g Mass Percent  100 Mass of Solution, g Mass of Solute  Mass of Solvent  Mass of Solution www.ThesisScientist.comExample 13.1—Calculate the Mass Percent of a Solution Containing 27.5 g of Ethanol in 175 mL H O. 2 Given: 27.5 27.5 g g e etha thanol, 175 mL nol, 202.5 g solut H Oion 2 Find: by by mass mass Solution Map: mL g Et H OH, g O H O g H O g sol’n 2 2 2 1.00 g H O g solute 2 by Mass  100 g solute  g solvent  g solution 1 mL H O g solution 2 Relationships: 1 mL H O = 1.00 g 2 Solve: 1.00 g H O 27.5 g ethanol 2 175 mL H O175 g H O by M2ass   2100 1 mL H O 2 202.5 g solution 27.5 g ethanol  175 g H O  202.5 g solution 2  13.6 The answer seems reasonable as it is less than 100. Check: www.ThesisScientist.comPractice—Calculate the Mass Percent of a Solution that Has 10.0 g of I Dissolved in 150.0 g of Ethanol. 2 www.ThesisScientist.comPractice—Calculate the Mass Percent of a Solution that Has 10.0 g of I Dissolved in 150.0 g of Ethanol, 2 Continued Given: 10.0 10.0 g g I I , , 160.0 g 150.0 g e solut thanol ionH O 2 2 2 Find: by by mass mass Solution Map: g EtOH, g H O g sol’n 2 g solute by Mass  100 g solute  g solvent  g solution g solution Relationships: Solve: 10.0 g I 150.0 g ethanol  160.0 g solution 10.0 g I 2 2 by Mass  100 160.0 g solution  6.25 The answer seems reasonable as it is less than 100. Check: www.ThesisScientist.comUsing Concentrations as Conversion Factors • Concentrations show the relationship between the amount of solute and the amount of solvent.  12 by mass sugar (aq) means 12 g sugar  100 g solution. • The concentration can then be used to convert the amount of solute into the amount of solution or visa versa. www.ThesisScientist.comExample 13.2—What Volume of 11.5 by Mass Soda Contains 85.2 g of Sucrose Given: 85.2 g sugar Find: volume, mL Solution Map: g solute g sol’n mL sol’n 100 g sol'n 1 mL sol'n 11.5 g sucrose 1.00 g sol'n Relationships: 100 g sol’n = 11.5 g sugar, 1 mL solution = 1.00 g Solve: 100 g 1 mL 85.2 g sugar 741 mL 11.5 g sugar 1.00 g The unit is correct. The magnitude seems reasonable Check: as the mass of sugar  10 the volume of solution. www.ThesisScientist.comPractice—Milk Is 4.5 by Mass Lactose. Determine the Mass of Lactose in 175 g of Milk, Continued. Given: 175 g milk  175 g solution Find: g lactose Equivalence: 4.5 g lactose  100 g solution Solution Map: g solution g Lactose 4.5 g Lactose 100 g solution Apply Solution Map: 4.5 g Lactose 175 g solution 7.9 g Lactose 100 g solution Check Answer: Units are correct. Number makes sense because lactose is a component of the mixture, therefore, its amount should be less. www.ThesisScientist.comPreparing a Solution • Need to know amount of solution and concentration of solution. • Calculate the mass of solute needed. Start with amount of solution. Use concentration as a conversion factor. 5 by mass 5 g solute  100 g solution. “Dissolve the grams of solute in enough solvent to total the total amount of solution.” www.ThesisScientist.comExample—How Would You Prepare 250.0 g of 5.00 by Mass Glucose Solution (Normal Glucose) Given: 250.0 g solution Find: g glucose Equivalence: 5.00 g glucose  100 g solution Solution Map: g solution g glucose 5.00 g Glucose 100 g solution Apply Solution Map: 5.00 g glucose 250.0 g solution 12.5 g glucose 100 g solution Answer: Dissolve 12.5 g of glucose in enough water to total 250.0 g. www.ThesisScientist.comSolution Concentration Molarity • Moles of solute per 1 liter of solution. • Used because it describes how many molecules of solute in each liter of solution. • If a sugar solution concentration is 2.0 M , 1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar: moles of solute Molarity = liters of solution www.ThesisScientist.comPreparing a 1.00 M NaCl Solution Weigh out Add water to 1 mole (58.45 g) dissolve the of NaCl and add NaCl, then it to a 1.00 L add water to Swirl to mix. volumetric flask. the mark. Step 1 Step 2 Step 3 www.ThesisScientist.comExample 13.3—Calculate the Molarity of a Solution Made by Dissolving 15.5 g of NaCl in 1.50 L of Solution Given: 15.5 g NaCl, 1.50 L solution Find: M Solution Map: g NaCl mol NaCl M L solution mol M L Relationships: M = mol/L, 1 mol NaCl = 58.44 g Solve: 0.2652 mol NaCl M 1 mol NaCl 15.5 g NaCl 0.2652 mol NaCl 1.50 L 58.44 g NaCl M  0.177 M Check: The unit is correct, the magnitude is reasonable. www.ThesisScientist.comUsing Concentrations as Conversion Factors • Concentrations show the relationship between the amount of solute and the amount of solvent.  0.12 M sugar (aq) means 0.12 mol sugar  1.0 L solution. • The concentration can then be used to convert the moles of solute into the liters of solution, or visa versa. • Since we normally measure the amount of solute in grams, we will need to convert between grams and moles. www.ThesisScientist.comExample 13.4—How Many Liters of a 0.114 M NaOH Solution Contains 1.24 mol of NaOH Given: 1.24 mol NaOH Find: volume, L Solution Map: mol NaOH L solution 1.00 L sol'n 0.114 mol NaOH Relationships: 1.00 L solution = 0.114 mol NaOH Solve: 1.00 L 1.24 mol NaOH10.9 L 0.114 mol NaOH The unit is correct, the magnitude seems reasonable as Check: the moles of NaOH 10x the amount in 1 L. www.ThesisScientist.comExample—How Would You Prepare 250 mL of 0.20 M NaCl Given: 250 mL solution Find: g NaCl Equivalence: 0.20 moles NaCl  1 L solution; 0.001 L = 1 mL; 58.44 g = 1 mol NaCl Solution Map: 0.20 mol NaCl mL L moles g 1 L solution 0.001 L solution NaCl NaCl 58.44 g 1 mL 1 mol NaCl Apply Solution Map: 0.001 L 0.20 mol NaCl 58.44 g 250 mL 2.9 g NaCl 1 mL 1L 1 mol NaCl Answer: Dissolve 2.9 g of NaCl in enough water to total 250 mL. www.ThesisScientist.comPractice—How Would You Prepare 100.0 mL of 0.100 M K SO (MM = 174.26) 2 4 www.ThesisScientist.comPractice—How Would You Prepare 100.0 mL of 0.100 M K SO (MM = 174.26), Continued 2 4 Given: 100.0 mL solution Find: g K SO 2 4 Equivalence: 0.100 moles K SO 1 L solution; 0.001 L = 1 mL; 2 4 174.26 g = 1 mol K SO 2 4 Solution map: 0.100 mol K SO 2 4 mL L moles g 1 L solution 0.001 L solution K SO K SO 2 4 2 4 174.26 g 1 mL 1 mol K SO 2 4 Apply solution map: 0.001 L 0.100 mol K SO 174.26 g 2 4 100.0 mL  1.74 g K SO 2 4 1mL 1 L 1 mol Answer: Dissolve 1.74 g of K SO in enough water to total 100.0 mL. 2 4 www.ThesisScientist.comMolarity and Dissociation • When strong electrolytes dissolve, all the solute particles dissociate into ions. • By knowing the formula of the compound and the molarity of the solution, it is easy to determine the molarity of the dissociated ions. Simply multiply the salt concentration by the number of ions. www.ThesisScientist.comMolarity and Dissociation + NaCl(aq) = Na (aq) + Cl (aq) 1 “molecule” = 1 ion + 1 ion 100 “molecules” = 100 ions + 100 ions 1 mole “molecules” = 1 mole ions + 1 mole ions + 1 M NaCl “molecules” = 1 M Na ions + 1 M Cl ions + 0.25 M NaCl = 0.25 M Na + 0.25 M Cl www.ThesisScientist.comMolarity and Dissociation, Continued 2+ CaCl (aq) = Ca (aq) + 2 Cl (aq) 2 1 “molecule” = 1 ion + 2 ion 100 “molecules” = 100 ions + 200 ions 1 mole “molecules” = 1 mole ions + 2 mole ions 2+ 1 M CaCl = 1 M Ca ions + 2 M Cl ions 2 2+ 0.25 M CaCl = 0.25 M Ca + 0.50 M Cl 2 www.ThesisScientist.comExample 13.5—Determine the Molarity of the Ions in a 0.150 M Na PO (aq) Solution. 3 4 Given: 0.150 M Na PO (aq) 3 4 + 3− Find: concentration of Na and PO , M 4 + 3− Na PO (aq)  3 Na (aq) + PO (aq) Relationships: 3 4 4 3 Solve: 1 mol PO 3 4 0.150 M Na PO 0.150 M PO 3 4 4 1 mol Na PO 3 4  3 mol Na  0.150 M Na PO 0.450 M Na 3 4 1 mol Na PO 3 4 The unit is correct, the magnitude seems reasonable as Check: the ion molarities are at least as large as the Na PO . 3 4 www.ThesisScientist.comPractice—Find the Molarity of All Ions in the Given Solutions of Strong Electrolytes. • 0.25 M MgBr (aq). 2 • 0.33 M Na CO (aq). 2 3 • 0.0750 M Fe (SO ) (aq). 2 4 3 www.ThesisScientist.comPractice—Find the Molarity of All Ions in the Given Solutions of Strong Electrolytes, Continued. 2+ • MgBr (aq) → Mg (aq) + 2 Br (aq) 2 0.25 M 0.25 M 0.50 M + 2 • Na CO (aq) → 2 Na (aq) + CO (aq) 2 3 3 0.33 M 0.66 M 0.33 M 3+ 2 • Fe (SO ) (aq) → 2 Fe (aq) + 3 SO (aq) 2 4 3 4 0.0750 M 0.150 M 0.225 M www.ThesisScientist.comDilution • Dilution is adding extra solvent to decrease the concentration of a solution. • The amount of solute stays the same, but the concentration decreases. • Dilution Formula: Conc x Vol = Conc x Vol start soln start soln final soln final sol • Concentrations and volumes can be most units as long as they are consistent. M1V1=M2V2 www.ThesisScientist.comExample—What Volume of 12.0 M KCl Is Needed to Make 5.00 L of 1.50 M KCl Solution Given: Initial solution Final solution Concentration 12.0 M 1.50 M Volume L 5.00 L Find: L of initial KCl Equation: (conc )∙(vol ) = (conc )∙(vol ) 1 1 2 2 concvol Rearrange and apply equation: 2 2 vol 1 conc 1 1.50 M5.00 L vol 1 12.0 M vol 0.625 L 1 www.ThesisScientist.comMaking a Solution by Dilution M x V = M x V 1 1 2 2 M = 12.0 M V = L 1 1 M = 1.50 M V = 5.00 L 2 2 M V M V 1 1 2 2 M V 2 2 V  1 M 1 1.50 M5.00 L V 0.625 L 1 12.0 M Dilute 0.625 L of 12.0 M solution to 5.00 L. www.ThesisScientist.comExample—Dilution Problems • What is the concentration of a solution made by diluting 15 mL of 5.0 sugar to 135 mL M = 5.0 M = 1 2 V = 15 mL V = 135 mL 1 2 (5.0)(15 mL) = M x (135 mL) 2 M = 0.55 2 • How would you prepare 200 mL of 0.25 M NaCl solution from a 2.0 M solution (2.0 M) x V = (0.25 M)(200 mL) M = 2.0 M M = 0.25 M 1 2 1 V = mL V = 200 mL V = 25 mL 1 2 1 Dilute 25 mL of 2.0 M NaCl solution to 200 mL. www.ThesisScientist.comPractice—Determine the Concentration of the Following Solutions. • Made by diluting 125 mL of 0.80 M HCl to 500 mL. • Made by adding 200 mL of water to 800 mL of 400 ppm. www.ThesisScientist.comPractice—Determine the Concentration of the Following Solutions, Continued. • Made by diluting 125 mL of 0.80 M HCl to 500 mL. M = 0.80 M M = M 1 2 V = 125 mL V = 500 mL 1 2 (0.80 M)(125 mL) = M x (500 mL) 2 M = 0.20 M 2 • Made by adding 200 mL of water to 800 mL of 400 ppm. M = 400 ppm M = ppm 1 2 V = 800 mL V = 200 + 800 mL 1 2 (400 PPM)(800 mL) = M x (1000 mL) 2 M = 320 PPM 2 www.ThesisScientist.comExample—To What Volume Should You Dilute 0.200 L of 15.0 M NaOH to Make 3.00 M NaOH V = 0.200L, M = 15.0 M, M = 3.00 M • Sort Given: 1 1 2 information. Find: V , L 2 • Strategize. Solution Map: V , M , M V 1 1 2 2 MV 1 1 V 2 M 2 Relationships: M V = M V 1 1 2 2 mol • Follow the Solve:  15.0 0.200 L  solution map L  1.00 L to Solve the mol  3.00  problem. L  Since the solution is diluted by a factor • Check. Check: of 5, the volume should increase by a factor of 5, and it does. www.ThesisScientist.comPractice Question 1—How Would You Prepare 400 mL of a 4.0 Solution From a 12 Solution www.ThesisScientist.comPractice Question 1—How Would You Prepare 400 ML of a 4.0 Solution From a 12 Solution, Continued M = 12 M = 4.0 (12) x V = (4.0)(400 mL) 1 2 1 V = mL V = 400 mL V = 133 mL 1 2 1 Dilute 133 mL of 12 solution to 400 mL. www.ThesisScientist.comPractice Question 2—How Would You Prepare 250 mL of a 3.0 Solution From a 7.5 Solution www.ThesisScientist.comPractice Question 2—How Would You Prepare 250 ML of a 3.0 Solution From a 7.5 Solution, Continued M = 7.5 M = 3.0 (7.5) x V = (3.0)(250 mL) 1 2 1 V = mL V = 250 mL V = 100 mL 1 2 1 Dilute 100 mL of 7.5 solution to 250 mL. www.ThesisScientist.comSolution Stoichiometry • We know that the balanced chemical equation tells us the relationship between moles of reactants and products in a reaction. 2 H (g) + O (g) → 2 H O(l) implies that for every 2 2 2 2 moles of H you use, you need 1 mole of O and will 2 2 make 2 moles of H O. 2 • Since molarity is the relationship between moles of solute and liters of solution, we can now measure the moles of a material in a reaction in solution by knowing its molarity and volume. www.ThesisScientist.comExample 13.7—How Many Liters of 0.115 M KI Is Needed to React with 0.104 L of a 0.225 M Pb(NO ) 3 2 2 KI(aq) + Pb(NO ) (aq) 2 KNO (aq) + PbI (s) 3 2 3 2 Given: 0.104 L Pb(NO ) 3 2 Find: L KI Solution Map: L mol mol L Pb(NO ) Pb(NO ) KI KI 3 2 3 2 1 L 2 mol KI 0.225 mol 0.115 mol 1 mol Pb(NO ) 3 2 1 L 0.225 mol Pb(NO ) = 1 L; 2 mol KI = 1 mol Pb(NO ) Relationships: 3 2 3 2; 0.115 mol KI = 1 L Solve: 0.225 mol Pb(NO ) 2 mol KI 1 L 3 2 0.104 L Pb(NO ) 0.407 L KI 3 2 1 L 1 mol Pb(NO ) 0.115 mol KI 3 2 The unit is correct. Check: www.ThesisScientist.comWhy Do We Do That • We spread salt on icy roads and walkways to melt the ice. • We add antifreeze to car radiators to prevent the water from boiling or freezing.  Antifreeze is mainly ethylene glycol. • When we add solutes to water, it changes the freezing point and boiling point of the water. www.ThesisScientist.comColligative Properties • The properties of the solution are different from the properties of the solvent. • Any property of a solution whose value depends only on the number of dissolved solute particles is called a colligative property. It does not depend on what the solute particle is. • The freezing point, boiling point, and osmotic pressure of a solution are colligative properties. www.ThesisScientist.comSolution Concentration Molality, m • Moles of solute per 1 kilogram of solvent. Defined in terms of amount of solvent, not solution. Does not vary with temperature. Because based on masses, not volumes. mole of solute molality  kg of solvent Mass of solution = volume of solution x density. Mass of solution = mass of solute + mass of solvent. www.ThesisScientist.comExample 13.8—What Is the Molality of a Solution Prepared by Mixing 17.2 g of C H O with 2 6 2 0.500 kg of H O 2 Given: 17.2 g C H O , 0.500 kg H O 2 6 2 2 Find: m mol Concept Plan: g C H O mol C H O 2 6 2 2 6 2 m kg m kg H O 2 Relationships: m = mol/kg, 1 mol C H O = 62.07 g 2 6 2 1 mol C H O Solve: 0.2771 mol C H O 2 6 2 2 6 2 17.2 g C H O 0.2771 mol 2 6 2 m 62.07 g C H O 2 6 2 0.500 kg H O 2 m  0.554 M Check: The unit is correct, the magnitude is reasonable. www.ThesisScientist.comFreezing Points of Solutions • The freezing point of a solution is always lower than the freezing point of a pure solvent.  Freezing point depression. • The difference between the freezing points of the solution and pure solvent is directly proportional to the molal concentration. •DT = m x K f f  K = freezing point constant. f • Used to determine molar mass of compounds. www.ThesisScientist.comFreezing and Boiling Point Constants Solvent K FP K BP f b °C °C C/m C/m Water, H O 1.86 0.00 0.512 100.0 2 Benzene, C H 5.12 5.53 2.53 80.1 6 6 Cyclohexane,C H 20.0 6.47 2.79 80.7 6 12 Naphthalene, C H 6.9 80.2 5.65 218 10 8 Ethanol, C H OH 1.99 115 1.22 78.4 2 5 tbutanol, (CH ) COH 8.3 25.6 82.4 3 3 Carbon tetrachloride,CCl 29.8 22.3 5.02 76.8 4 Methanol, CH OH 97.8 0.80 64.7 3 Acetic acid, HC H O 3.9 16.7 3.07 118 2 3 2 www.ThesisScientist.comExample 13.9—What Is the Freezing Point of a 1.7 m Aqueous Ethylene Glycol Solution, C H O 2 6 2 1.7 m C H O (aq) Given: 2 6 2 Find: T , °C f Solution Map: mDT FP f DT m K FP − FP = DT solv sol’n f f Relationships:DT = m ∙K , K for H O = 1.86 °C/m, FP = 0.00 °C f f f 2 H2O Solve: DT m K FP FPDT f f ,H O H O soln ' f 2 2 C 0.00C FP 3.2C   1.7 m 1.86 soln ' m FP3.2C soln ' DT 3.2 C f Check: The unit is correct, the freezing point being lower than the normal freezing point makes sense. www.ThesisScientist.comBoiling Points of Solutions • The boiling point of a solution is always higher than the boiling point of a pure solvent. Boiling point elevation. • The difference between the boiling points of the solution and pure solvent is directly proportional to the molal concentration. •DT = m x K b b K = boiling point constant. b www.ThesisScientist.comExample 13.10—What Is the Boiling Point of a 1.7m Aqueous Ethylene Glycol Solution, C H O 2 6 2 1.7 m C H O (aq) Given: 2 6 2 Find: T , °C b Solution Map: mDT BP b DT m K BP − BP = DT b b sol’n solv DT = m ∙K , K H O = 0.512 °C/m, BP = 100.00 °C Relationships: b b b 2 H2O Solve: DT m K BP BPDT b b,H O solution solvent b 2 C BP100.00C 0.87C solution 1.7 m0.512 m BP100.87C sol'n DT 0.87 C f Check: The unit is correct, the boiling point being higher than the normal boiling point makes sense. www.ThesisScientist.comPractice—What Is the Boiling Point of a Solution that Has 0.20 moles of Sulfur Dissolved in 0.10 kg of Cyclohexane (BP = 80.7 C, K = 2.79 C/m) cyclohexane b www.ThesisScientist.comPractice—What Is the Boiling Point of a Solution that Has 0.20 moles of Sulfur Dissolved in 0.10 kg of Cyclohexane, Continued 0.20 mol S, 0.10 kg cyclohexane Given: Find: T , °C b Solution Map: mol S, kg solvent mDT BP b mol DT m K BP − BP = DT m sol’n solv b b kg Relationships:DT = m ∙K , K = 2.79 °C/m, BP =80.7 °C, m = mol/kg b b b Solve: DT m K BP BPDT b b solution cyclohexane b 0.20 mol S C m BP 80.7C 5.58C   2.0 m 2.79 solution 0.10 kg m BP 86.3C solution  2.0 m DT 5.58 C b Check: The unit is correct, the boiling point being higher than the normal boiling point makes sense. www.ThesisScientist.comOsmosis and Osmotic Pressure • Osmosis is the process in which solvent molecules pass through a semipermeable membrane that does not allow solute particles to pass.  Solvent flows to try to equalize concentration of solute on both sides.  Solvent flows from side of low concentration to high concentration. • Osmotic pressure is pressure that is needed to prevent osmotic flow of solvent. • Isotonic, hypotonic, and hypertonic solutions.  Hemolysis. www.ThesisScientist.comDrinking Seawater Because seawater has a higher salt concentration than your cells, water flows out of your cells into the seawater to try to decrease its salt concentration. The net result is that, instead of quenching your thirst, you become dehydrated. www.ThesisScientist.comOsmotic Pressure Solvent flows through a semipermeable membrane to make the solution concentration equal on both sides of the membrane. The pressure required to stop this process is osmotic pressure. www.ThesisScientist.comHemolysis and Crenation Normal red blood Red blood cell in Red blood cell in cell in an isotonic a hypotonic hypertonic Solution. solution. solution. Water Water flows into flows out the cell, eventually of the cell, causing eventually causing the cell to burst. the cell to distort and shrink. www.ThesisScientist.com