GENERAL CHEMISTRY CHEM 162
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This 25 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 162 at University of Washington taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/192605/chem-162-university-of-washington in Chemistry at University of Washington.
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Date Created: 09/09/15
Chapter 17 Properties of Solutions 176 177 178 BoilingPoint Elevation and Freezing Point Depression Osmotic Pressure Colligative Properties of Electrolyte Solutions Colloids De nitions for Solutions Solute The smaller in mass of the components in a solution the material dispersed into a solvent Solvent The major component of the solution the material that the solute is dissolved into Solubility The maximum amount that can be dissolved into a particular solvent to form a stable solution at a speci ed temperature Miscibility The ability of two substances to dissolve in one another in any proportion The formation of solutions from pure substances is always favored by ENTROPY When two pure substances are mixed the disorder of the system is in creased As with any chemical reaction solvation is a balance between enthalpy and entropy N it Gig In 3322 MW Glbbs Free Energy E f if E x xquot E IILIJ AG AH TAS 37qu Cl faq In ows hum 39 Wmul A A quantitative description of solvation The Solution Czcle Step 1 Solute separates into particles overcoming attractions Therefore Endothermic Step 2 Solvent separates into particles overcoming intermolecular attractions Therefore Endothermic solvent aggregated heat Igt solvent separated A H gt o solvent Step 3 Solute and Solvent particles miX particles attract each other Therefore Exothermic solute separated solvent separated Igt solution heat Hlt0 A The Thermochemical Cycle AH AH solution 2 solute AHsolvent A1 1mix If AHEndothemnc steps lt AHExothenmc steps solut1on becomes warmer If AH gt AH solution becomes colder Endothermic steps Exothermic steps Solution Cycles and the Enthalpy Components of the Heat of Solution Solute Solvent led separated I 7 I mg 4 i cga g Hsolum EL E F39 H 39VEi t advent 4 Salute E Eagle a l egated Mm LL 7 Emisth see Fig 172 nal I I39m it Sa vent quotSalute m I separated V 5 12quot Hsulum Sggginn Ea 3391 QH gi 39 Holman a I M E I I I I I F Sullluemt m Salute 1 a 7 ggr ga ed g aglgm egaied M I Msalin 393 a n B Envdnthermic snlutinn prince55 Himiitial What happens when solutes dissolve Iondlpole Intermolecular if Forces in H bond 39 Methanol QH O Solutions CHM Dipoledipole CHDH chloroform Ioninduced dipole 73 cHCIa cl Hexane csHm Dipoleinduced dipole G H20 Xenon Dispersion Xe Octane 55H caHm Chl7 Motto Like Dissolves Like Polar molecules dissolve best in Polar solvents Polar molecules can exhibit strong dipoledipole intermolecular interactions with polar solvents hence increasing their solubility Nonpolar molecules dissolve best in non polar solvents To dissolve a nonpolar molecule in a polar solvent requires breaking strong dipoledipole interactions of the solvent Key Point Forces created between solute and solvent must be comparable in strength to the forces destroyed within both Like dissolves Like solubility of methanol in water I FF E Methanol A solution of water and melhanol Table 131 Solubility of Alcohols in Water and llexane Alcohol Model Water Hexane CH 3OIl 399 methanol a m 03912 CHSCHEOH I w on ethanol CH3CH220H no no propanol CH3CH23OH h utanol 03911 m CH3CH OH pentazn ol 03903 CH3CH250H hexanol 039005398 m Expressed in mol aloohol 00 gl solvent at 20 C Hydration 7shells Hydration shells around an aqueous ion h AH mewm a Enthampy AH 39 a H 31 gum 3 DHag warns exane e m u g separated 33 5 E 4 w E Mao and j Hexane NEGHS 4 aggregated aggregated Hexame A it I Hexane Octane ET separated separated AHmlm E H Manatee 5 e E i H 3 AHme Lu I l exane I I V ctene I d aggregated 1 aggregated Hmma lSolutionl InHmn 0 E nal Predicting Relative Solubilities of Substances Problem Predict which solvent will dissolve more of the given solute a Sodium Chloride in methanol CH3OH or in propanol CH3CH2CH20H b Ethylene glycol HOCHZCHZOH in water or in hexane CH3CH2CH2CH2CH2CH3 c Diethyl ether CH3CH20CH2CH3 in ethanol CH3CH20H or in water Plan Examine each solute and solvent to determine which intermolecular forces will be active A solute tends to be more soluble in a solvent that has the same type of intermolecular forces active Predicting Relative Solubilities of Substances Solution a Methanol NaCl is an ionic compound that dissolves through ion dipole forces Both methanol and propanol contain a polar OH group but propanol s longer hydrocarbon chain would interact only weakly with the ions and be less effective in stabilizing the ions b Water Ethylene glycol has two OH groups and is stabilized by extensive H bonding in water c Ethanol Diethyl ether shows both dipolar and dispersion intermolecular forces and could form H bonds to both water and ethanol The ether would be more soluble in ethanol because solvation in water must disrupt many more strong H bonding interactions ile 134 Concentration Definitions Concentration Term Ratio M I t M amount moi oi soiute D a H voiuma L oi solution Molality m amount mol oi soiute mass kg oi soivent mass of solute Parts Iby mass mass of solution volume oi soiute volume of solution amount moi oi soiute Parts by uoiume Mole iraotion X amount moi of soiute amount moi oi soivent Remember Concentrations are ratios They are not additive Volumes are additive Vtota1 V1 V2 V3 Masses are additive mtota1 In1 m2 m3 Moles are additive ntota1 n1 n2 n3 Concentrations are not additive 0X01 c2 c3 Calculating Molalig I To calculate Molality molkg we need the number of moles of solute and the mass of solvent used to dissolve the solute Normally we are given the mass of solute and mass of solvent therefore we calculate the moles of the solute from the mass then use the mass of the solvent to calculat Remember Molality is different from Molarity Molality is based on mass and is independent of temperature or pressure unlike molarity divide by kg water Because 1 L of H20 welghs 1 kg molality and molarity of dilute aquous solutions are nearly identical Calculating Molality 11 Problem Determine the molality and molarity of a solution prepared by dissolving 750g BaNO32 S in to 37400g of water at 250C Plan We convert the quantity of BaNO32 to moles using the molar mass and then divide by the volume of H20 in liters using water density 099707 gml3 Solution molar mass of BaNO32 26132 gmol 750 moles Ba 0 L 028700 1 N 392 26132 gmol m0 6 molality W 076739 m 0767 m 037400 kg molarity we need the volume of solution and can assume that addition of the salt did not change the total volume 37400 H O 375099 ml 03750991 099707 W M 2 W 0765 M 03750991 Expressing Concentrations in Parts by Mass Problem Calculate the Parts per Billion ppb by mass of iron in a 185 g Fe supplement pill that contains 00543 ug of Fe Plan Convert ug Fe to grams and then use Fe mass pill and multiply by 109 to obtain ppb Solution 00543 ug Fe 543 X 10 39 8 g Fe 8 5431220 Fe x 109 294 Parts per Billi0n ppb g Expressing Concentrations in Parts by Volume Problem The label on a can of beer 340 ml indicates 45 alcohol by volume What is the volume in liters of alcohol it contains Plan we know the vol and the total volume so we use the de nition to nd the volume of alcohol Solution 4 5 ml alcohol 1 Al h l 39 2 V0 co 0 100 m1 beer X 340 ml beer 153 ml alcohol Expressing Concentrations in Mole Fraction Problem A sample of alcohol contains 118g of ethanol CZHSOH and 37 50g of water What are the mole fractions of each Plan We know the mass and formula of each compound so we convert both to moles and apply the de nition of mole fraction Solution Moles Ethanol 118 EthanOI 2565 mol Ethanol 46 g Ethanolmol Moles Water 395 H O 2194 mol H20 18 g HZOmol XEthanol 2565 010467 2194 2565 mer AL 089533 2194 2565 Converting Concentration Units Problem Commercial concentrated Hydrochloric acid is 118 M HC1 and has a density of 1190 gml Calculate the a mass HC1 b molality and 0 mole fraction of HC1 Plan We know Molarity and density a For mass HC1 we need the mass of HC1 and water the solvent Assume 1L of solution from the density we know the mass of the solution and from the molecular mass of HC1 we calculate its mass b We know moles of HC1 and mass of water c we use moles HC1 froma and use the mass of water to get moles of water then calculate mole fractions and add them to check Solution a assume 1L of HC1 solution 118 moles HC1 118 moles HC1 x 3646 HC1 4302 g HC1 mole HC1 Converting Concentration Units II a cont 1000 mL 1190 g soln 1 L solution mL soln 1L solution X 1190 g solution mass HC1430392 HC1 x 100 3615 HCl 1190 g solutlon b mass of H20 mass of solution mass of HCl 1190 g solution 4302 g HCl 7597 g H20 118 moles HCl 1553 HCl 07597 kg H20 m 6 7597 g H70 4217 1 H 0 18016 g HZOmol H20 0 e 2 Total moles 42172 118 53972 540 Solution XHC1 0219 XH20 42 17 0781 40 540
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