1、1Chemistry I2 Chapters 7&823Parts of a Solution SOLUTE the part of a solution that is being dissolved(usually the lesser amount)SOLVENT the part of a solution that dissolves the solute(usually the greater amount)Solute+Solvent=SolutionSoluteSolventExamplesolidsolidMetal alloyssolidliquidKool aidliqu
2、idliquidAlcoholic drinksgasliquidPepsigasgasair452.Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.6Miscibility Miscible:compounds that dissolve readily in each other in any proportion.Ex:water and alcohol.Immiscible:Liquids that do not readily d
3、issolve in each other Ex:oil and water7Solubility Solubility is relative,most substances will dissolve by extremely small amounts.Soluble:the solute will dissolve in a particular solvent greater than 1 g/100 mL Insoluble:the solute will dissolve only 0.1 g/100 mL,or not dissolve in the solvent891011
4、.Molarity(M)=moles soluteliters of solution12135.00 g 1 mol237.7 g=0.0210 mol0.0210 mol0.250 L=0.0841 M=0.0841 M1415Practice For homework:Pg 242#1-6,9 Pg 268#19&20 (Read pg 266-268)16Solutions(continued)Day 2Mrs.Kay17Learning CheckHow many grams of NaOH are required to prepare 400.mL of 3.0 M NaOH s
5、olution?1)12 g2)48 g3)300 g1819m of solution=mol solutekilograms solvent2021conc(molality)=1.00 mol glycol0.250 kg H2O 4.00 molal%glycol =62.1 g62.1 g+250.g x 100%=19.9%22Learning CheckA solution contains 15 g Na2CO3 and 235 g of H2O?What is the mass%of the solution?1)15%Na2CO32)6.4%Na2CO33)6.0%Na2C
6、O323Using mass%How many grams of NaCl are needed to prepare 250 g of a 10.0%(by mass)NaCl solution?24Try this molality problem 25.0 g of NaCl is dissolved in 5000.mL of water.Find the molality(m)of the resulting solution.m=mol solute/kg solvent25 g NaCl 1 mol NaCl 58.5 g NaCl=0.427 mol NaClSince the
7、 density of water is 1 g/mL,5000 mL=5000 g,which is 5 kg0.427 mol NaCl 5 kg water=0.0854 molal salt water2526Practice for homework:Read pg 255-263 Do Pg 261#5-8 (%m/m)Do Pg 268#19-20(molarity)27Changing Concentration:Our calculations for this unit are based on solutions being carefully made with kno
8、wn concentrations.Two ways to have a known concentration:1.Dissolving a measured mass of pure solute in a certain volume of solution.(We practiced this as molarity questions)2.Diluting a solution of known concentration to the new concentration youre trying to achieve.28Diluting a Standard solution:U
9、se the formula:C1v1=C2v2You will be given 3 of the 4 variables and need to solve for the unknown.C1=the original concentrated solutionv1=the amount of original concentrated solution used.C2=the new diluted concentration of solutionv2=the new volume of diluted solution29Practice:I have 2.0 L of 0.10
10、M of sulfuric acid.This usually sold as an 18 M concentrated solution.How much of the concentrated solution do I need to make my new solution?YOU KNOW:C1=18 M;C2=0.10 M;v2=2.0 LSo,v1=C2v2C1 v1=(0.10 M)(2.0 L)(18 M)v1=0.011 L or 11 mL30More Practice:Page 273#25-27 Finish the Handout called“Molarity R
11、eview”#5-931CHEM 11ASolution StoichiometryYou can solve for amounts of product made or reactants needed even though they are stated as mol/L rather than g or moles.You need to use the basics that you learned in stoichiometry from grade 11 chemistry and apply it to what youve learned about solutionsL
12、ets work through an example32How many grams of Ca(NO3)2 can be prepared by reacting 75 mL of 2.6 mol/L HNO3 with an excess of Ca(OH)2?2 HNO3(l)+Ca(OH)2(s)Ca(NO3)2(aq)+2H2O(l)What are you given?V HNO3=0.075L and M=2.6 mol/LFind the moles of HNO3 used:2.6 mol/L x 0.075 L=0.195 mol3.Use stoichiometry t
13、o find the ratio of acid moles to calcium nitrate.0.195 mol(acid)x 1 mol(calcium nitrate)=0.0975 mol 2 mol(acid)332 HNO3+Ca(OH)2 Ca(NO3)2+2H2O4.0.0975 mol calcium nitrate needs to be converted to mass.0.0975 mol Ca(NO3)2 X 164.04 g =15.99 g mol Ca(NO3)2 So,we will make 15.99 g of calcium nitrate during this reaction.34Practice:Handout#56-60
