C CHEM 102 1st Edition Lecture 1 Outline of Last Lectture I No N previouss Lecture Outline of Current Lecture II Chapter C 14 Solutions aand their Behavior Conccepts a Basics of solutio ons and solvvents b Rulees of solubility c Conccentration units u d System Thermodynamics e State function i Enthalpyy iii Entropyy iiii Free Eneergy Currentt Lecture a Basics B of solu utions and solvents Solution A homo ogenous mixxture of two o or more subsstances moleecules n or in otheer words it iss the compon nent Solveent The majjority speciess in a solution present in the larrgest amount usually a liiquid ority species in a solution that could eeither originaate as a liquid d gas or Solute The mino d and it dissolves in the so olution solid Example given in claass When ad dding salt to water the saalt is the solu ute since it iss the a originatees as a solid species s and the t water smaller sspecies beinggs added to tthe solution and is the sollvent since itt is the majorrity species in n the solution Together tthe solute an nd the solvent m make a solution SOLUTIO ON SALT WA ATER SOLU UTE SALT SOLVENT W WATER Hom mogenous mixture A sam mple of matteer that appeaars to be the same througghout It may consist of a single chemiical substancce or a mixture of differen nt solutions As mixed referrenced in class it is a mixxture that is ccompletely m b Rules R of Solu ubility Thesee rules are ussed in order tto determinee if a particular solute will dissolve w when it is added to a particular solvent A general rule of thum mb is that waater is usuallyy the s solvent Two o rules were eemphasized in lecture Rulee 1 All ionic ssalts involving alkali metaal cations eleements in column 1 and the amm monium ion HN4 are higghly soluble As a brief reminder solu uble means th hat it is able to be dissolvved oride is solub ble as it breaks down into o Na which is a alkali Exaample NaCl sodium chlo metaal cation and d Cl an anion n Rulee 2 All nitratee and acetatee salts are veery soluble These are the diiagrams for the molecular make ups ple Ca NO3 2 and Pb acettate 2 are solu uble when pllaced in wateer Examp c Co oncentration n Units Molaarity the M symbol is commonly used to represent molarity It represent tthe moles of so olute per liter of solution Molarity usses the volum me Equaation M of i moles of o i Litter of solu ution i is the solute mmonly used d to represen nt molality Itt represents the solute Molaality the m ssymbol is com per kkilogram of ssolvent i Molality uses th he weight or mass Equaation m of i moles of o i K of solvent Kg Examplee Ethylene gglycol EG haas a Moleculaar Weight of 62 1 grams m mol 1 2 kg of EG was added tto 4 0 kg of w water The resulting solutiion had a volume of 4 1 LL Find m and M ore being ablee to find m yyou need to ffind the molees of EG To ffind the moles you Befo take the weight ggiven of the EG E and conveert the kg intto grams and d divide by th he moleecular weightt of the EG given nEGG 1 2kg 1000 1 g 1mo ole 19 3 moles EG G 1 kg 62 1g Now w molality can be calculatted using thee moles of the EG kg of the t solvent w which is the wateer m 19 3 moles EG 4 m 4 8 4 0 kG H 2O To find the molarity you take the moles of the Eg L of the solution M 19 3 moles EG 4 7 M 4 1 L solution Mole Fraction symbol chi moles of species i divided by the sum of the moles of all i species in a solution The sum of the mole fractions must be 1 Equation i ni ni n j Example 1 2 kg of EG Molecular Weight 62 1 grams mole was added to 4 0 kg of water Fine the mole fraction of EG in the solution In the earlier example we found out that there were 19 3 moles of EG by taking the 1 2 kg of EG dividing it by 1000 and dividing that by the molecular weight of 62 1 grams For the moles of water you divide the grams of water by the molecular weight 4000 g H 2O 1 mole nH 2O 222 mole H 2O 1 18 g H 2O To find the mole ratio of EG you take the moles of the EG over the total moles the moles of EG and the moles of water 19 3 moles EG EG 0 080 19 3 moles EG 222 moles H 2O Percent by weight i w w mass of species i divided by the sum of masses of all species in solution multiplied by100 Equation Wti i w w 100 Wti Wt j Example Using the 1 2 kg of EG and the 4 0 kg of water Find the percent weight of EG For this you take the weight of the EG and divide it by the total weight the kg of the water and the kg of the water and multiply it by 100 to get a percentage wt EG w w kg EG 1 2 100 100 23 kgEG kgH2O 1 2 4 0 So this means that there are 23 grams of EG per 100 grams of solution Percent by volume i v v volume of species i divided by the sum of volumes of all species in a solution multiplied by 100 Equation Vi i v v 100 Vi V j TRICK FOR CONVERSIONS Density is the key to convert from one unit to the other This can be used when having to find the mass or volume of a species You can use the density mass volume to solve for a certain variable that you are looking for Here is one example where density would be used when we are converting molality to molarity Example Concentrated nitric acid has a molarity of 16 M The density of the solution is 1 42 g mL Find the molality So you are given 16M which you know is 16 moles of HNO3 over 1 liter of solution since that is the definition of molarity and we are trying to find the molality which you know is the number of moles per kg of solvent In this case we are going to assume that the volume is 1 Liter of solution because we know that there are 16 moles of HNO3 per 1 liter of solution We want to know the number of kg of H2O in 1 liter of solution since we are usually going to use water as the solvent in most cases in this class The following steps are taken to find the grams of solution the nitric acid and the water 1 liter …