Note: the precision of the instruments must be used to ensure all concentrations are reported to the correct number of significant figures and with units Part 1: Preparing Stock Solutions a) Prepare 200.0mL of a 0.500MNaCl solution in distilled water Calculations: C=Vnn=cV Mole of NacL =0.1 volum =200mL. molerity =0.500m moler =1000concentrationxvolume =10000.500200=0.1 Moler mass of Nacl= 23.99+35.5=58.5g/ mal Actual Mass of NaCl=5.85g All equipment and glassware needed to make this solution: weighing.Pan, small Beaker Scoopula, taploading Balance, volumetric flask, parafilm, funnl. Exact concentration of the final NaCl solution: C=vnM= Mole Nacl / Volume of Solution =MolarmassNaclMassNacl= of M=(5.8500g/58.5g/Mol)/0.2000L =0.5000M Calculations: Mole of KI=10000.125100=0.0125 moler Moler Mass of KI=39+127=166g/mol Mass of KI=0.0125166=2.075g Actual Mass of KI=2.075g All equipment and glassware needed to make this solution: Beaker, weighing, Volumetric Flask, analytical batance spatula. Exact concentration of the final Kl solution: Volume of solution =100.0mL or 0.1000L M=(2.0750g/166g/mol)/0.10000L=0.1200r Solution approved by instructor (instructor initials): Part 2: Diluting Stock Solutions From the stock 1.21MHCl solution provided, prepare the two solutions listed below. Focus on obtaining a concentration as close to the desired value as possible, using the most accurate analytical glassware available. a) 50.00mL of a 0.570MHCl solution ==Calculations:c1v1=c2v250.000.570=1.21v2v2=1.2150.000.570=v2=23.5ML Actual Volume of HCl stock solution =25.00ML Exact concentration of the final working HCl solution: Solution approved by instructor (instructor initials): b) 250.0mL of a 0.0500MHCl solution Calculations: C1=0.0500V2= ? Calculations:C1C1V1=C2V20.050250=1.21V2=0.0500V2=?C2=1.21V1=250mLV2=1.210.0500250V2=10.3mL Actual Volume of HCl stock solution = All equipment and glassware needed to make this solution: Exact concentration of the final working HCl solution: Solution approved by instructor (instructor initials): Note: the precision of the instruments muat be used to ensuee all cencentrationsare reported to the cerrect ewenter of signticant figures and wath units. Rart ti: Breparing Stock Solutiens a) Prepare 200.9mL. of a 0.500MNaCl selhation in distilled water. Calculations cmVNn=Cx,V500m male of Nacl =0.1 volum=200mL.Molerity=0.500m10000.500200=0+1AatsofNacl=0.159.5=5.8.59 Achat Mass of NaCl = 5.355 A ll equipnint and glassware feeded to make this solfion: Woighing Pan , smat 5sok6x scoopula. toploading Balance, volumbtric Flask, Parafim, Funnl. m=(5.5500g(58.5g/mel)/0.2000L=0.500m Actual Mass of ki =2.075z Al equipment and glassware needed so make this solution. Beaker, welghing, Volumetric Flask. aneelytical batance spatula Exact concentration of the final KI solusion. Velame of solution = leo.o mL or o. oroe L M=(2.0750g/166s/Nel)/s1oveel=0.12oM Solution approved by instruclor (nstructor initais) Part 2: Diluting 5tock Solutions From the stock 1.21MHCi solusion provided, prepare the two solulices listed below. Focus on obtaining a concencration as dose to the desired value as possitie, using the most accurate analytical glassware avalable. a) 50.00mL of a 0.570MHCl solution Calculationsc1V1=c2V2=50.000.570=1.21V2.=V2=1.2150.000.570=V2=2.5.ML Actuat Volume of HCi stock solution = 25.00ML Exact concentration of the final working HCl solution: Solution approved by instructor (instructor initials): b) 250.0mL of a 0.0500MHCl solution Calculations:C1=0.0500V2=?C1V1=C2V20.050250=1.21V2V2=0.0500250C2=1.21V1=250MLV2=10.3ML Actual Volume of HCl stock solution = All equipment and glassware needed to make this solution: Exact concentration of the final working HCl solution: Solution approved by instructor (instructor initials)