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Solution FormationSolutions are homogeneous mixtures that maybe solid, liquid, or gaseous.The compositions of the solvent and the solute determine whether a substance will dissolve.Stirring (agitation), temperature, and the surface area of the dissolving particles determine how fast the substance will dissolve.These three factors involve the contact of the solute with the solvent.
Stirring & Solution FormationStirring speeds up the process of dissolving because fresh solvent is continually brought into contact with the surface of the soluteStirring affects only the rate at which a solid solute dissolves. It does not influence the amount of solute that will dissolve.An insoluble substance remains undissolved regardless of how vigorously or for how long the solvent/solute system is agitated.
Temperature & Solution FormationAt higher temperatures, the kinetic energy of the solvent molecules is greater than at lower temperatures so they move faster.The more rapid motion of the solvent molecules leads to an increase in the frequency and the force of the collisions between the solvent molecules and the surfaces of the solute molecules.
Particle Size & Solution FormationA spoonful of granulated sugar dissolves more quickly than a sugar cube because the smaller particles in granulated sugar expose a much greater surface area to the colliding solvent molecules.The more surface of the solute that is exposed, the faster the rate of dissolving.
SolubilityIf you add 36.0 g of NaCl to 100 g H2O at 25ºC, all of the 36.0 g of salt dissolves.If you add one more gram of salt and stir (no matter how vigorously or for how long) only 0.2 g of the last portion will dissolve.According to the kinetic theory, water molecules are in continuous motion.They should continue to bombard the excess solid, removing and solvating the ions.
SolubilityAs ions are solvated, they dissolve in the water.Based on this information, you might expect all of the salt to dissolve eventually. That does not happen, however, because an exchange process is occurring.New particles from the solid are solvated and enter into solution. At the same time an equal number of already dissolved particles crystallize.These particles come out of solution and are deposited as a solid. The mass of undissolved crystals remains constant.
SolubilityIn a saturated solution, a state of dynamic equilibrium exists between the solution and the excess solute.The rate of solvation (dissolving) equals the rate of crystallization, so the total amount of dissolved solute remains constant.The system will remain the same as long as the temperature remains constant.Saturated solution – contains the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure.
SolubilityExample: 36.2 g of salt dissolved in 100 g of water is a saturated solution at 25ºC.If additional solute is added to this solution, it will not dissolve.Solubility of a substance is the amount of solute that dissolves in a given quantity of a solvent at a specified temperature and pressure to produce a saturated solution.Solubility is often expressed in grams of solute per 100 g solvent. (gas sometimes g/L)
SolubilityUnsaturated solution – a solution that contains less solute than a saturated solution at a given temperature and pressure.If additional solute is added to an unsaturated solution, it will dissolve until the solution is saturated.Some liquids are infinitely soluble in each other. Any amount will dissolve in a given volume.Two liquids are miscible if they dissolve in each other in all proportions (water and ethanol)
Factors Affecting SolubilityTemperature affects the solubility of a solid, liquid and gaseous solutes in a solvent.Both temperature and pressure affect the solubility of gaseous solutes.The solubility of most solid substances increases as the temperature of the solvent increases.Mineral deposits form around the edges of hot springs because the hot water is saturated with minerals. As the water cools, some of the minerals crystallize because they are less soluble at the lower temperature.
Factors Affecting SolubilityFor a few substances, solubility decreases with temperature.Supersaturated solution – contains more solute than it can theoretically hold at a given temperature.Make a saturated solution of sodium acetate at 30·C and let the solution stand undisturbed as it cools to 25ºC.You would expect that solid sodium acetate will crystallize from the solution as the temperature drops. But no crystals form.
Supersaturated SolutionsThe crystallization of a supersaturated solution can be initiated if a very small crystal, called a seed crystal, of the solute is added.Rock candy is another example of crystallization in a supersaturated solution.A solution is supersaturated with sugar and seed crystals cause the sugar to crystallize out of solution.A supersaturated solution crystallized rapidly when disturbed.
Temperature and Gas SolubilityThe solubilities of most gases are greater in cold water than in hot.Thermal pollution happens when an industrial plant takes water from a lake for cooling and then dumps the heated water back into the lake.The temperature of the lake increases which lowers the concentration of dissolved oxygen in the lake water affecting aquatic animal and plant life.
Pressure and SolubilityChanges in pressure have little affect on the solubility of solids and liquids, but pressure strongly influences the solubility of gases.Carbonated beverages contain large amounts of carbon dioxide dissolved in water. Dissolved CO2 makes the drink fizz.The drinks are bottle under higher pressure of CO 2 gas, which forces large amounts of the gas into solution.When opened, the partial pressure of CO2 above the liquid decreases.
Pressure and SolubilityImmediately, bubbles of CO2 form in the liquid and escape from the bottle and the concentration of dissolved CO2 decrease.If the drink is left open, it becomes “flat” as it loses its CO2.Henry’s Law – sated that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid.As the pressure of the gas above the liquid increases, the solubility of the gas increases.
Pressure and Solubility Henry’s Law S1 = S2 P1 P2
QuestionThe solubility of a gas in water is 0.16 g/L at 104 kPa. What is the solubility when the pressure of the gas ins increased to 288 kPa. Assume the temperature remains constant. S1 = S2 P1 P2 (288 kPa) ( 0.16g/L) = 4.4 x 10-1 g/L (104 kPa)
QuestionsWhat determines whether a substance will dissolve?Chemical composition of the solute and solvent.What determines how fast a substance will dissolve?Agitation, temperature and particle size of the soluteWhat units are usually used to express the solubility of a solute?g of solute per 100 g solvent
QuestionsWhat would you do to change a saturated solid/liquid solution to an unsaturated solution?Add solventWhat would you do to change a saturated gas/liquid solution to an unsaturated solution?Increase the pressureWhat are two conditions that determine the mass of solute that will dissolve in a given mass of solvent?Temperature and pressure (if the solute is a gas)
ConcentrationConcentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent.Dilute solution is one that contains a small amount of solute.Concentrated solution – contains a large amount of solute.In chemistry the most important unit of concentration is molarity.
MolarityMolarity (M) is the number of moles of solute dissolved in one liter of solution Molarity (M) = moles of solute / liters of solution.Note that the volume involved is the total volume of the resulting solution, not the volume of the solvent alone.3 M NaCl is read as “three molar sodium chloride”
Molarity QuestionsA solution has a volume of 2.0 L and contains 36.0 g of glucose (C6H12O6). If the molar mass of glucose is 180 g/mol, what is the molarity of the solution?M = moles of solute / L of solutionM = 36.0 g glucose 1 mol glucose 180 g glucose 2.0 LM = 0.1mol/L or 0.1M C6H12O6
Molarity QuestionsA solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity? M = moles of solute / L of solution M = 0.70 mol NaCl 0.250L solution (convert to L) M = 2.8 mol/L or 2.8M NaCl
Molarity QuestionsHow many moles are in each quantity of the following substances? 12.0 g NaCl (12.0 g NaCl) (1 mole NaCl / 58.5 g NaCl) = 0.205 mole NaCL 53.8 g KNO3 (53.8 g KNO3) ( 1 mol KNO3 / 101.1g KNO3) = 0.532 mol KNO3
Molarity QuestionsFind the mass in grams of each of these amounts of substances? 1.5 mol NaOH (1.5 mol NaOH) (40 g NaOH / 1 mol NaOH) = 60 g of NaOH 0.575 mol NaHCO3(0.575 mol NaHCO3)( 84g NaHCO3 / 1 mol NaHCO3) 48.3 g NaHCO3
Molarity QuestionsWhich contains more molecules 1.00 mol SO2 or 1.00 mol SO3? They both contain the same number of moleculesWhich contains more mass 1.00 mol SO2 or 1.00 mol 1 mole of SO3 (80.1g) – 1 mole of SO2 (64.1g)
MolaritySometimes you may need to determine the number of moles of solute dissolved in a given volume of solution. How many moles are in 2.00 L of 2.5M lithium chloride (LiCl)? Moles of solute = molarity (M) x liters of solution (V) Moles of solute = (2.5 moles/L) ( 2.00L) Moles of solute = 5.0 mol
Molarity QuestionsHow many moles of ammonium nitrate are in 335 mL of 0.425 M NH4NO3? mol NH4NO3 = M x L of solution = (0.425 mol/L) (0.335L) = 0.142 mol NH4NO3How many moles of solute are in 250 mL of 2.0M CaCl2? How many grams of CaCl2 is that?mol CaCl2 = (2.0 mole/L) (0.250L) = 0.50 mol0.50 mol CaCl2 (111.11 g CaCl2 / 1 mol CaCl2 ) = 56 g
Making DilutionsDiluting - To make less concentrated by adding solvent.Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change.Moles of solute before dilution = moles of solute after dilutionmoles of solute = M x L of solution and total number of moles of solute remains unchanged upon dilution. M1V1 = M2V2
Making Dilutions M1V1 = M2V2 molarity & volume molarity and volume of original solution of diluted solutionVolumes can be L or mL as long as the same units are used for both V1 and V2
Making DilutionsA student is preparing a 100 mL of 0.40M MgSO 4 from a stock solution of 2.0 M MgSO4. How would she do this? M1V1 / M2 = V2V2 = 20 ml – She would measure 20 mL of the stock solution (2.0 M MgSO4) and transfer it to a volumetric flask.Then she would add water to the flask to make 100 mL of solution.Try the class activity on page 482
QuestionsHow many milliliters of a solution of 4.0 M KI are needed to prepare a 250.0 mL of 0.760 M KI?V1 = (0.760M)(250.0 mL) / (4.0 M) = 47.5 mLHow could you prepare 250 mL of 0.20M NaCl using on a solution of 1.0M NaCl and water?V1 = (0.20M) ( 250 mL) / ( 1.0 M) = 50 mLUse a pipet to transfer 50 mL of the 1.0M solution to a 250 mL flask. Then add distilled water up to the mark.
Percent Solutions (v / v)The concentration of a solution in percent can be expressed in two ways:As the ratio of the volume of the solute to the volume of the solution or as the ratio of the mass of the solute to the mass of the solutionPercent by volume (% (v/v)) = volume of solute x 100% volume of solutionHow many milliliters of isopropyl alcohol are in 100 mL of 91% alcohol?
QuestionIf 10mL of acetone (C3H6O) is diluted with water to a total solution volume of 200mL, what is the percent by volume of acetone in the solution?Percent by volume (% (v/v)) = volume of solute x 100% volume of solution% by volume of acetone = 10 mL / 200 mL =5.0% v/v
QuestionA bottle of the antiseptic hydrogen peroxide is labeled 3.0% (v/v). How many mL hydrogen peroxide are in a 400.0 mL bottle of this solution?Percent by volume (% (v/v)) = volume of solute x 100% volume of solution0.03 = x mL / 400.0 mL(0.03) (400.0 mL) = x12 mL = x
Percent Solutions (mass/mass)Another way to express the concentration of a solution is as a percent (mass/mass), which is the number of grams of solute in 100 g of solution.A solution containing 7 g of NaCl in 100 g of solution is 7% (m/m)Percent by mass (% (m/m) = mass of solute x 100% mass of solution
Percent Solutions (mass/mass)You want to make 2000g of a solution of glucose in water that has a 2.8% (m/m) concentration of glucose. How much glucose should you use?Percent by mass (% (m/m) = mass of solute x 100% mass of solution2000 g solution=(2.8g glucose/100 g solution) = 56 g glucoseHow much solvent should be used? The mass of the solvent equals the mass of the solution minus the mass of the solute.(2000 g – 56 g ) = 1944 g of solventThus a 2.8% (m/m) glucose solution contains 56 g of glucose dissolved in 1944 g of water.
QuestionsHow do you calculate the molarity of a solution?Molarity (M) = moles of solute / volume of solutionCompare the number of moles of solute before dilution with the number of moles of solute after dilution.The total number of moles of solute in solution does not change (only the number of moles of solute per unit volume)What are two ways of expressing the concentration of a solution as a percent? Volume solute Mass of solute Volume of solution Mass of solution
Colligative Properties of SolutionsThe physical properties of a solution differ from those of the pure solvent used to make the solution.Some of these differences in properties have little to dowith the specific identity of the solute.They depend upon the number of solute particles in the solution.Colligative Property – a property that depends only upon the number of solute particles, and not upon their identity.
Colligative Properties of SolutionsThree important colligative properties of solutions are:1. Vapor pressure lowering2. Boiling point elevation3. Freezing point depressionVapor pressure is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system.A solution that contains a solute that is nonvolatile (not easily vaporized) always has a lower vapor pressure than the pure solvent.
Colligative Properties of SolutionsIn a pure solvent, equilibrium is established between the liquid and the vapor.In a solution, solute particles reduce the number of free solvent particles able to escape the liquid. Equilibrium is established at a lower vapor pressure.Ionic solutes that dissociate, such as sodium chloride and calcium chloride, have greater effects on the vapor pressure than does a non-dissociating solute such as glucose.Each formula unit of CaCl2 produces three particles in solution, a calcium ion and two chloride ions.
Colligative Properties of SolutionsThe decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution.3 moles of NaCl dissolved in H2O produce 6 mol of particles - each formula unit dissociates into 2 ions3 moles of CaCl2 dissolved in H2O produce 9 mol of particles - each formula unit dissociated into 3 ions3 moles of glucose dissolved in water produce 3 mol of particles – glucose does not dissociate.
Colligative Properties of SolutionsThe vapor pressure lowering caused by 0.1 mol of NaCl in 1000 g of water is twice that caused by 0.1 mol of glucose in the same quantity of water.The vapor pressure lowering caused by 0.1 mol of CaCl2 in 1000 g of water is three times that caused by 0.1 mol of glucose in the same quantity of water.The decrease in a solution’s vapor pressure is proportional to the number of particles the solute
Freezing-Point DepressionWhen a substance freezes, the particles of the solid take on an orderly pattern.The presence of a solute in water disrupts the formation of this pattern because of the shells of water of solvation. (water molecules surround the ions of the solute)As a result, more KE must be withdrawn from a solution than from the pure solvent to cause the solution to solidify.The freezing point of a solution is lower than the freezing point of the pure solvent.
Freezing-Point DepressionFreezing-Point Depression – the difference in temperature between the freezing point of a solution and the freezing point of the pure solvent.Freezing-point depression is another colligative property.The magnitude of the freezing-point depression is proportional to the number of solute particles dissolved in the solvent and does not depend upon their identity.The addition of 1 mol of solute particles to 1000 g of water lowers the freezing point by 1.86ºC.
Freezing-Point DepressionIf you add 1 mole (180g) of glucose to 1000 g of water, the solution freezes at -1.86ºC.If you add 1 mol (58.5g) of NaCl to 1000 g of water, the solution freezes at -3.72ºC, double the change for glucose.This is because 1 mol NaCl produces 2mol particles and doubles the freezingpoint depression.Salting icy surfaces forms a solution withthe melted ice that has a lower freezingpoint than water. (antifreeze also)
RemindersIonic compounds and certain molecular compounds, such as HCl, produce two or more particles when they dissolve in water.Most molecular compounds, such as glucose, do not dissociate when they dissolve in water.Colligative properties do not depend on the kind of particles, but on their concentration.Which produces a greater change in colligative properties – an ionic solid or a molecular solid?An ionic solid produces a greater change because it will produce 2 or more mole of ions for every mol of solid that dissolves.
Boiling-Point ElevationBoiling Point – of a substance is the temperature at which the vapor pressure of the liquid phase equals atmospheric pressure.Adding a nonvolatile solute to a liquid solvent decreases the vapor pressure of the solvent.Because of the decrease in vapor pressure, additional KE must be added to raise the vapor pressure of the liquid phase of the solution to atmospheric pressure and initiate boiling.Thus the boiling point of a solution is higher than the boiling point of the pure solvent.
Boiling-Point ElevationBoiling Point Elevation – The difference in temperature between the boiling point of a solution and the boiling point of the pure solvent.The same antifreeze, added to automobile engines to prevent freeze-ups in winter, protects the engine from boiling over in summer.Boiling-point elevation is a colligative property, it depends on the concentration of particles, not on their identity.It takes additional KE for the solvent particles to overcome the attractive forces that keep them in the liquid.
Boiling-Point ElevationThe magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in the solvent.The boiling point of water increases by 0.512ºC for every mole of particles that the solute forms when dissolved in 1000g of water.To make fudge, a lot of sugar and some flavoring are mixed with water and the solution is boiled. As the water slowly boils away, the concentration of sugar in the solution increases.As the concentration increases, the boiling point steadily rises.
QuestionsWhat are three colligative properties of solutions?Vapor-pressure lowering, boiling-point elevation, and freezing-point depressionWhat factor determines how much the vapor pressure, freezing point, and boiling point of a solution differ from those properties of the pure solvent?The number of solute particles dissolved in the solvent.Would a dilute or a concentrated sodium fluoride solution have a higher boiling point? Explain.Concentrated NaF, - magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in the solvent.
QuestionsAn equal number of moles of KI and MgF2 are dissolved in equal volumes of water. Which solution has the higher boiling point? MgF2 Higher vapor pressure? KI solution Higher freezing point? KI solution
Molality and Mole FractionUnit molality and mole fractions are two additional ways in which chemists express the concentration of a solution.Molality (m) is the number of moles of solute dissolved in 1 kg of solvent. Molality (m) = moles of solute / kg of solvent Molarity = moles of solute / L of solutionIn the case of water as the solvent, 1 kg = 1000 mL, 1000 g = 1 L
MolalityTo prepare a solution that is 1.00 molal (1m) in glucose, you add 1 mol (180g) of glucose to 1000g of water.0.500 molal solution in sodium chloride is prepared by dissolving 0.50 mol (29.3 g) of NaCl in 1.0 kg of waterMolality (m) = moles of solute / kg of solventThe molality of a solution does not wary with temperature because the mass of the solvent does not change.Molarity = moles of solute / L of solutionThe molarity of a solution does vary with temperature because the liquid can expand and contract.
Molality QuestionsHow many grams of NF are need to prepare a 0.400mNaF solution that contains 750g water? 750 g H2O 0.400 mol NaF 42g NaF = 13g NaF 1000 g H2O mol NaFCalculate the molality of a solution prepared by dissolving10.0g of NaCl in 600 g of water.10.0 g NaCl 1 mol NaCl 1000 g H2O = 2.85 x 10-1m 600g H 2O 58.5 g NaCl 1 kg H2O
Mole FractionThe concentration of a solution also can be expressed as a mole fraction.Mole fraction of a solute in a solution is the ratio of the moles of the solute to the total number o moles of solvent and solute.In a solution containing nA mole of solute A and nB mole of solvent B, the mole fraction of solute A and the mole fraction of solvent B can be expressed as follows. XA = nA XB = nB nA + nB nA + n B
Mole Fraction QuestionsWhat is the mole fraction of each component in a solution made by mixing 300 g of ethanol (C2H5OH) and 500 g water. Xethanol = nethanol nethanol + nwater300 g C2H5OH 1 mol C2H5OH = 6.7 mol C2H5OH 45 g C2H5OH500 g H2O 1 mol H2O = 27.8 mol H2O 18 g H2O Xethanol = 6.7 mol C2H5OH = 0.190 m C2H5OH
Mole Fraction QuestionsWhat is the mole fraction of each component in a solution made by mixing 300 g of ethanol (C2H5OH) and 500 g water. Xwater = nwater nethanol + nwater300 g C2H5OH 1 mol C2H5OH = 6.7 mol C2H5OH 45 g C2H5OH500 g H2O 1 mol H2O = 27.8 H2O 18 g H2O Xwater = 27.8 mol C2H5OH = 0.810 H2O
Mole Fraction QuestionsCalculate the mole fraction of each component in a solution of 42g CH3OH, 35g C2H5OH, and 50 g C3H7OH XA = nA nA + nB + n C42 g CH3OH 1 mol CH3OH = 1.3 mol CH3OH 32 g CH3OH35 g C2H5OH 1 mol C2H5OH = 0.76 mol C2H5OH 46 g C2H5OH50 g C3H7OH 1 mol C3H7OH = 0.83 mol C3H7OH 60 g C3H7OH
Freezing Point Depression and Boiling Point ElevationThe freezing point of a solvent is lowered and its boiling pointis raised by the addition of a nonvolatile solute.The magnitudes of the freezing point depression (ΔTf) andthe boiling point elevation (ΔTb) of a solution are directlyproportional to the molal concentration (m) when the solute ismolecular, not ionic. ΔTb Ñ m ΔTf Ñ mChange in the freezing temperature (ΔTf ) is the differencebetween the fp of the solution and the fp of the pure solvent.Change in boiling temperature (ΔTb ) is the differencebetween the bp of the solution and the bp of the pure solvent.
Molal Freezing Point Depression ConstantWith the addition of a constant, the proportionality betweenthe ΔTf and the molality (m) can be expressed in an equation ΔTf = Kf x mThe constant, Kf, is the molal freezing-point depressionconstant, which is equal to the change in freezing point for aconstant1 molal solution of a nonvolatile molecular solute.The value of Kf depends upon the solvent. Its units are ºC/m.
Molal Boiling Point Elevation ConstantThe boiling-point elevation of a solution can also beexpressed as an equation ΔTb = Kb x mThe constant, Kb, is the molal boiling-point elevationconstant, which is equal to the change in boiling point for a 1constantmolal solution of a nonvolatile molecular solute.The value of Kb depends upon the solvent. Its units are ºC/m.For ionic compounds, both the freezing point depression andthe boiling point elevation depend upon the number of ionsproduced by each formula unit
ProblemsWhat is the freezing point depression (and boiling pointelevation) of an aqueous solution of 10.0 g of glucose(C6H12O6) in 50.0 g H2O?10.0 g C6H12O6 1 mol = 0.0555 mol C6H12O6 180 g m = mol solute = 0.055 mol = 1.11 m kg solvent .0500 kgΔTf = Kf x m = (1.86 ºC/m) (1.11m) = 2.06 ºCΔTb = Kb x m = (0.512 ºC/m) (1.11 m) = 0.568 ºC
ProblemsCalculate the freezing point depression of a benzenesolution containing 400 g of benzene and 200 g of themolecular compound acetone (C3H6O). Kf for benzene is5.12 ºC/m200 g C3H6O 1 mol = 3.45 mol C3H6O 58 g m = mol solute = 3.45 mol = 8.63 m kg solvent .400 kgΔTf = Kf x m = (5.12 ºC/m) (8.63m) = 44.2 ºC
ProblemsFreezing points and boiling points cannot be depressed orelevated without end.As the concentration of a solute increases, there comes apoint when the quantity of the solute exceeds the quantityof the solvent.At this point, the solute then becomes the solvent becauseis present is a higher concentration.
QuestionWhich solution has a higher boiling point, 1 mol of Al(NO 3)3in 1000g of water or 1.5 mol of KCl in 1000 g of water?The solution of Al(NO3)3 has a higher boiling point becauseAl(NO3)3 dissociates into a larger number of particles.Why is it important to distinguish between nonvolatile andvolatile compounds when discussing certain colligativeproperties?Volatile solutes would quickly evaporate at highertemperatures, which would change the molal concentrationof the solution.
Calculating the Boiling Point of an Ionic SolutionWhat is the boiling point of a 1.5m NaCl solution?Each formula unit of NaCl dissociates into two particles,Na+ and Cl-, the effective molality is 2 x 1.5m = 3.00m.Calculate the boiling point elevation and then add it to100ºC.ΔTb = Kb x m = (0.512 ºC/m) (3.00m) = 1.54 ºC Boiling Point = 100ºC + 1.54ºC = 101.54ºC
Ionic SolutionsWhat is the boiling point of a solution that contains 1.25mol CaCl2 in 1400 g of water.ΔTb = Kb x m = (0.512 ºC/m) (2.68m) = 1.37 ºC Boiling Point = 100ºC + 1.37ºC = 101.37ºCWhat mass of NaCl would have to be dissolved in 1.000kg of water to raise the boiling point by 2.00ºCΔTb = Kb x m = (0.512 ºC/m) (?m) = 2.00 ºC m = 3.91 / 2 = 1.96 (Na+ and Cl-)1.96 mol NaCl 58.5 g = 115 g NaCl 1 kg solution 1 mol
QuestionsWhat are two ways of expressing the ratio of soluteparticles to solvent particles?Molality and mole fractionsHow are freezing point depression and boiling pointelevation related to molality?The magnitudes of the freezing point depression and theboiling point elevation of a solution are directlyproportional to the molal concentration when the solute ismolecular, not ionic.
QuestionsHow many grams of sodium bromide must be dissolvedin 400.0 g of water to produce a 0.500 molal solution?0.500 mol 102.9 g .4000 kg = 20.6 g NaBr1 kg molCalculate the mole fraction of each component in a solutionof 2.50 mol ethanoic acid in 10.00 mol of water.Xethanoic acid = 2.5 mol / 12.5 mol = 0.200X water = 10.0 mol / 12.5 mol = 0.800
QuestionsWhat is the freezing point of a solution of 12.0 g of CCl4dissolved in 750.0 g of benzene? The freezing point ofbenzene is 5.48 ºC; Kf is 5.12 ºC/mm = 12.0 g CCl4 1 mol = 0.104 m 154 g 0.7500kgΔKf = m x Kf = (0.104m) ( 5.12 ºC/m) = 0.53ºCFreezing point = 5.48ºC – 0.53ºC = 4.95ºC