(Solution Manual) Fundamentals of Analytical Chemistry

Skoog, West, Holler, Crouch, 9

Chemistry

Solution Manual

9th Edition

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Solution Manual for Fundamentals of Analytical Chemistry, The downloadable file consists of 646 pages, and the total size of the file is 11.91 MB. Purchase now only for highly discounted price.

Author(s): Skoog, West, Holler, Crouch

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Fundamentals of Analytical Chemistry: 9th ed. Chapter 3 Chapter 3 3-1. (a) SQRT returns the square root of a number or result of a calculation. (b) AVERAGE returns the arithmetic mean of a series of numbers. (c) PI returns the value of pi accurate to 15 digits (d) FACT returns the factorial of a number, equal to 1  2  3  …  number. (e) EXP returns e raised to the value of a given number. (f) LOG returns the logarithm of a number to a base specified by the user. 3-2. Count(value 1, value2,…) returns the number of cells that contain numbers and numbers within the list of arguments. As it appears in Figure 3 -10, the COUNT function should return a value of 8 for the number of data values in each column of the spreadsheet. 3-3. One method is to add comments for the appropriate c ells. The comment group is found under the Review tab. An example for finding the mean (average) of four numbers is shown below. Another method is to add a text box to the worksheet by clicking on Text Box in the Insert tab. An arrow could be added pointi ng to the appropriate cell by clicking on Shapes in the Illustrations group under the Insert tab. Fundamentals of Analytical Chemistry: 9th ed. Chapter 3 3-4. Replacing these values gives the worksheet below. Obviously, Column C has a nonsensical result with the replacement. 3-5. The result is: 3-6. The answer is contained in the problem. 2 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 Chapter 4 4-1. (a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule or an electron. There are 6.021023 particles  10−3 mole = 6.02 1020 particles mole millimole millimole (b) The molar mass is the mass in grams of one mole of a chemical species. (c) The millimolar mass is the mass in grams of one millimole of a chemical species. (d) Parts per million, cppm, is a term expressing the concentration of dilute solutions. Thus, cppm = mass of solute mass of solution  106 ppm The units of mass in the numerator and the denominator must be the same. 4-2. The molar species concentration is number of moles of that species contained in one liter of solution. The molar analytical concentration is the total number of moles of a solute in 1 liter of the solution, regardless of the solute’s chemic al state. 1000 mL 1 cm3  1 m 3 4-3. The liter: 1 L =    = 10−3 m3 1 L 1 mL  100 cm  Molar concentration: 1 M = 1 mol  1 L = 1 mol 1 L 10−3 m3 10−3 m3 4-4. (a) 3.2 108 Hz  1 MHz 106 Hz = 320 MHz −7 109 ng (b) 4.56 10 g  = 1 g 456 ng Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 2 3 4 2 4 (c) 8.43107 μmol  1 mol 106 μmol = 84.3 mol (d) 6.51010 s  1 Gs 109 s = 65 Gs (e) 8.96 106 nm  mm 106 nm = 8.96 mm (f) 48, 000 g  kg 1000 g = 48 kg 4-5. For oxygen, for example 15.999 u/atom = 15.999 g/6.022 ×1023 atoms = 15.999 g/mol. So 1 u = 1 g/mol. Thus, 1g = 1mol u. 4-6. From Pb. 4-5, 1 g = 1 mol u = 6.022 × 1023 u 1 u = 1/12 mass of 12C atom So 1 kg = 1000 g = 1000× Avogadro’s number of u = 1000 × Avogadro’s number × 1/12 mass of 12C atom = 1000/12 × Avogadro’s number × mass of 12C atom 1 mol Na PO 3 mol Na+ 6.022 1023 Na+ 22 + 4-7. 2.92 g Na3PO 4  3 4   = 3.22 163.94 g 1 mol Na PO 1 mol Na+  10 Na 4-8. 2 mol K+ 3.41 mol K2HPO4  1 mol K HPO  6.022 1023 K+ 1 mol K+ = 4.111024 K+ 4-9. (a) 8.75 g B O  2 mol B  1 mol B2O3 = 0.251 mol B 2 3 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 3 1 mol B2O3 69.62 g B2O3 167.2 mg Na B O •10H O 1 g  7 mol O 2 4 7 2 1000 mg 1 mol Na B O •10H O (b) 1 mol Na2B4O7 •10H2O = 3.07 381.37 g 2 4 7 2  10−3 mol O = 3.07 mmol Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 4 (c) 4.96 g Mn O  1 mol Mn3O4  3 mol Mn = 6.50  10−2 mol Mn 228.81 g Mn3O4 1 mol Mn3O4 333 mg CaC O  1 g  mol CaC2O4  2 mol C = 5.20  10−3 mol C (d) 2 4 = 5.20 mmol 1000 mg 128.10 g CaC2O4 1 mol CaC2O4 850 mg P O × 1 g × 1 mol P2O5  1000 mmol  2 mol P 4-10. (a) 1000 mg 141.94 g P2O5 1 mol 1 mol P2O5 = 11.98 mmol P (b) 40.0 g CO × 1 mol CO 2  1000 mmol  1 mol C = 909 mmol C 44.01 g CO2 1 mol 1 mol CO2 (c) 12.92 g NaHCO × 1 mol NaHCO3  1000 mmol  3 mol O = 461.4 mmol O 84.01 g NaHCO3 1 mol 1 mol NaHCO3 57 mg MgNH PO × 1 g × 1 mol MgNH4PO4 ×1000 mmol (d)  1000 mg 137.32 g MgNH4PO4 1 mol 1 mol Mg = 0.42 mmol Mg 1 mol MgNH 4PO 4 4-11. (a) 0.0555 mol KMnO 4 × 1000 mmol × 2.00 L = 111 mmol KMnO (b) L 1 mol 4 3.25  10−3 M KSCN 1000 mmol L × × ×750 mL L 1 mol 1000 mL = 2.44 mmol KSCN 3.33 mg CuSO4 × 1 g × 1 mol CuSO4 ×1000 mmol × 3.50 L 3 4 2 5 2 3 4 4 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 5 (c) 1 L 1000 mg 159.61 g CuSO 4 = 7.30×10−2mmol CuSO 1 mol (d) 0.414 mol KCl ×1000 mmol × 1 L  250 mL = 103.5 mmol KCl 1 L 1 mol 1000 mL 4 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 6 4 2 3 6 0.320 mol HClO4 ×1000 mmol × 1 L ×226 mL 4-12. (a) 1 L 1 mol 1000 mL = 72.3 mmol HClO 4 (b) 8.05×10−3mol K CrO 1000 mmol 2 4 × ×25.0 L 1 L 1 mol = 201 mmol K2CrO 4 6.75 mg AgNO3 × 1 g × 1 mol AgNO3 ×1000 mmol ×6.00 L (c) 1 L 1000 mg 169.87 g AgNO 3 = 0.238 mmol AgNO 3 1 mol 0.0200 mol KOH ×1000 mmol × 1 L ×537 mL (d) 1 L 1 mol 1000 mL = 10.7 mmol KOH 4-13. (a) 0.367 mol HNO × 63.01 g HNO3 ×1000 mg = 2.31 × 104mg HNO 1 mol HNO 3 1 g (b) 245 mmol MgO× 1 mol × 40.30 g MgO ×1000 mg = 9.87 × 103mg MgO 1000 mmol 1 mol MgO 1 g (c) 12.5 mol NH NO × 80.04 g NH4 NO3 ×1000 mg = 1.00 × 106mg NH NO 1 mol NH 4NO 3 1 g 4.95 mol (NH ) Ce(NO ) × 548.23 g (NH4 )2Ce(NO3 )6 ×1000 mg (d) 4 2 3 6 1 mol (NH4 )2Ce(NO3 )6 1 g = 2.71 × 106mg (NH ) Ce(NO ) 4-14. (a) 3.20 mol KBr × 119.0 g KBr 1 mol KBr = 381 g KBr 3 3 4 3 4 3 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 7 (b) 18.9 mmol PbO × 1 mol × 223.20 g PbO = 4.22 g PbO 1000 mmol 1 mol PbO Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 8 4 (c) 6.02 mol MgSO × 120.37 g MgSO4 = 725 g MgSO 4 1 mol MgSO 4 10.9 mmol Fe(NH ) (SO ) •6H O × 1 mol × 392.23 g Fe(NH4 )2 (SO4 )2 •6H2O (d) 4 2 4 2 2 = 4.28 g Fe(NH4 )2 (SO4 )2 •6H2O 1000 mmol 1 mol Fe(NH4 )2 (SO4 )2 •6H2O 0.350 mol sucrose × 1 L × 342 g sucrose ×1000 mg 4-15. (a) 1 L 1000 mL 1 mol sucrose 1 g × 16.0 mL = 1.92×103mg sucrose 3.76×10−3mol H O 34.02 g H O 1000 mg 2 2 × 2 2 × (b) 1 L 1 mol H2O2 1 g × 1.92 L = 246 mg H2O2 2.96 mg Pb(NO 3 )2 × 1 L × 356 mL (c) 1 L 1000 mL = 1.05 mg Pb(NO 3 )2 0.0819 mol KNO3 × 101.10 g KNO3 × 1000 mg × 1 L (d) 1 L 1 mol 1 g 1000 mL × 5.75 mL = 47.6 mg KNO 3 0.264 mol H2O2 × 1 L × 34.02 g H2O2 × 250 mL 4-16. (a) 1 L 1000 mL 1 mol H2O2 = 2.25 g H2O2 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 9 (b) 5.75×10−4mol benzoicacid 1 L 122 g benzoicacid × × 1 L 1000 mL 1 mol benzoicacid × 37.0 mL = 2.60×10−3g benzoicacid (c) 31.7 mg SnCl 2 × 1 g ×4.50 L = 0.143 g SnCl 1 L 1000 mg 2 0.0225 mol KBrO3 × 1 L ×167 g KBrO3 ×11.7 mL (d) 1 L 1000 mL 1 mol KBrO 3 = 4.40 × 10−2g KBrO 4-17. (a) pNa = − log(0.0635 + 0.0403) = − log(0.1038) = 0.9838 pCl = − log(0.0635) = 1.197 pOH = − log(0.0403) = 1.395 (b) pBa = −log(4.65 10−3 ) = 2.333 pMn = −log(2.54) = −0.405 pCl = −log(2  4.6510−3 + 2  2.54) = −log(5.089) = −0.707 (c) pH = −log(0.400) = 0.398 pCl = −log(0.400 + 2  0.100) = − log(0.600) = 0.222 pZn = −log(0.100) = 1.00 (d) pCu = −log(5.78 10−2 ) = 1.238 pZn = −log(0.204) = 0.690 3 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 10 pNO 3 = −log(2  0.0578 + 2  0.204) = − log(0.5236) = 0.281 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 11 6 4 (e) pK = −log(4 1.62 10−7 + 5.12 10−7 ) = −log(1.16 10−6 ) = 5.936 pOH = −log(5.12 10−7 ) = 6.291 pFe(CN) = −log(1.62 10−7 ) = 6.790 (f) pH = −log(4.75 10−4 ) = 3.32 pBa = −log(2.35 10−4 ) = 3.63 pClO = −log(2  2.3510−4 + 4.7510−4 ) = −log(9.45 10−4 ) = 3.02 4-18. (a) pH = 4.31, log[H 3O+] = −4.31, [H3O+] = 4.9 × 10–5 M as in part (a) (b) [H3O+] = 3.3 × 10−5 M (c) [H3O+] = 0.26 M (d) [H3O+] = 1.3 × 10−14 M (e) [H3O+] = 2.4 × 10−8 M (f) [H3O+] = 4.8 × 10−6 M (g) [H3O+] = 5.8 M (h) [H3O+] = 2.6 M 4-19. (a) pNa = pBr = −log(0.0300) = 1.523 (b) pBa = −log(0.0200) = 1.699; pBr = −log(2 × 0.0200) = 1.398 (c) pBa = −log(5.5 × 10–3) = 2.26; pOH = −log(2 × 5.5 × 10−3) = 1.96 (d) pH = −log(0.020) = 1.70: pNa = −log(0.010) = 2.00 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 12 3 pCl = −log(0.020 + 0.010) = −log(0.030) = 1.52 (e) pCa = −log(8.7 × 10−3) = 2.06; pBa = −log(6.6 × 10–3) = 2.18 pCl = −log(2 × 8.7 × 10−3 + 2 × 6.6 × 10−3) = −log(0.0306) = 1.51 (f) pZn = −log(2.8 × 10−8) = 7.55; pCd = −log(6.6 × 10−7) = 6.18 pNO 3 = −log(2.8 × 10−8 + 2 × 6.6 × 10−7) = 5.87 4-20. (a) pH = 1.020 ; log[H 3O+] = −1.020; [H 3O+] = 0.0955 M (b) pOH = 0.0025; log[OH–] = −0.0025; [OH–] = 0.99 M (c) pBr = 7.77; [Br–] = 1.70 × 10−8 M (d) pCa = −0.221; [Ca2+] = 1.66 M (e) pLi = 12.35; [Li+] = 4.5 × 10−13 M (f) pNO 3 = 0.034; [NO –] = 0.92 M (g) pMn = 0.135; [Mn2+] = 0.733 M (h) pCl = 9.67; [Cl–] = 2.14 × 10−10 M 3 + 1 1.02 g 1000 mL 1 mol Na+ −2 + 4-21. (a) 1.0810 ppm Na     = 4.79 10 M Na 106 ppm 1 mL 1 L 22.99 g 2− 1 1.02 g 1000 mL 1 mol SO 3− −3 2− 270 ppm SO 4     4 = 2.87 10 M SO 4 106 ppm 1 mL 1 L 96.06 g (b) pNa = −log(4.79 10−2 ) = 1.320 pSO4 = −log(2.87 10−3 ) = 2.542 4-22. (a) 300 nmol/L = 300 × 10−9 mol/L or 300 nM in plasma 2.2 mmol/L = 2.2 × 10−3 mol/L or 2.2 mM in whole blood Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 13 (b) pHb in plasma = −log(300 × 10–9) = 6.52 pHb in blood = −log(2.2 × 10−3) = 2.66 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 4-14. (a) 14 2 2 2 2 5.76 g KCl•MgCl2 •6H2O ×1 mol KCl•MgCl2 •6H2O =1.04×10−2M KCl•MgCl •6H O 2.00 L 277.85 g 2 2 (b) There is 1 mole of Mg2+ per mole of KCl MgCl2, so the molar concentration of Mg2+ is the same as the molar concentration of KCl MgCl2 or 1.04 × 10−2M −2 3 mol Cl− −2 − (c) 1.04×10 M KCl•MgCl2 •6H2O×1 mol KCl•MgCl •6H O =3.12×10 M Cl (d) 5.76 g KCl•MgCl2 •6H2O × 1 L ×100%=0.288% (w/v) 2.00 L 1000 mL 3.12 10−2 mol Cl− 1 L 1000 mmol −1 − (e)  × ×25 mL = 7.810 mmol Cl 1 L 1000 mL 1 mol −2 1 mol K+ 39.10 g K+ 1000 mg (f) 1.04 10 M KCl•MgCl2 •6H2O 1 mol KCl•MgCl •6H O   1 mol K+ 1 g = 407 mg 1 L = 407 ppm K+ (g) pMg = –log(1.04 × 10−2) = 1.983 (h) pCl = –log(3.12 × 10−2) = 1.506 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 15 3 6 3 3 mol Fe(CN) − 3 6 6 3 3 4-24. (a) 1210 mg K3Fe(CN)6  1000 mL  1 g = 1210 g K3Fe(CN)6 775 mL 1 L 1000 mg 775 L 1210 g K3Fe(CN)6 × mol K3Fe(CN)6 775 L 329.2 g = 4.74×10−3M K Fe(CN) (b) 4.74 10−3 3 mol K+ M K3 Fe(CN)6  mol K Fe(CN) = 1.42 10−2 M K+ (c) 4.74 10−3 M K Fe(CN) 3−  6 = 4.74 10 3 M Fe(CN) 3− mol K Fe(CN) 6 3 6 (d) 1210mgK3Fe(CN)6  775mL g 1000mg 100% = 0.156% (w/v) 1.42 10−2 mol K+ L 1000 mmolK+ −1 + (e)  × ×50 mL = 7.1×10 mmol K L 1000 mL mol K+ 4.74 10−3 mol K Fe(CN) mol Fe(CN) 3− 211.95 g Fe(CN) 3− (f) 3 6  6  6 L mol K Fe(CN) mol Fe(CN) 3− 1000 mg = 1005 mg Fe(CN) 3− = 1005 ppm Fe(CN) 3−  g L 6 6 (g) pK = –log(1.42 × 10–2) = 1.848 (h) pFe(CH) 6 = –log(4.74 × 10–3) = 2.324 4-25. (a) 6.42% Fe(NO 3 )3 = 6.42 g Fe(NO3 )3 ×1.059 g ×1000 mL ×1 mol Fe(NO3 )3 100 g solution mL 1 L 241.86 g = 2.8110−1 M Fe(NO ) = 0.281 M (b) − 2.81×10-1mol Fe(NO ) 3 mol NO − 6 6 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 16 2.8110 1M Fe(NO ) = 3 3  3 = 8.4310−1M NO − L 1 mol Fe(NO 3 )3 (c) 2.8110−1mol Fe(NO ) 241.86 g Fe(NO ) 3 3  3 3  1 L = 6.80 101g Fe(NO ) = 68.0 g L 1 mol 3 3 3 3 3 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 17 1.11 mol NiCl 2 mol Cl − 4-26. (a) 12.5% NiCl = 12.5 g NiCl2 ×1.149 g ×1000 mL ×1 mol NiCl2 =1.11 M NiCl 2 100 g solution 1 mL 1 L 129.61 g 2 − (b) 1.11 M NiCl2 = 2 × = 2.22 M Cl L 1 mol NiCl 2 (c) 1.11 mol NiCl2 ×129.61 g NiCl2 1 L = 1.44×102 g NiCl L mol 2 4-27. (a) 4.75 g C2H5OH  500 mL soln = 2.38101g C H OH 100 mL soln 2 5 Weigh 23.8 g ethanol and add enough water to give a final volume of 500 mL 4.75% (w/w) C H OH= 4.75 g C2H5OH  500 g soln = 2.38101g C H OH 2 5 100 g soln 2 5 (b) 500 g soln=23.8 g C2H5OH+x g water x g water = 500 g soln − 23.8 g C2H5OH = 476.2 g water Mix 23.8 g ethanol with 476.2 g water 4.75% (v/v) C H OH= 4.75 mL C2H5OH (c) 2 5 100 mL soln 4.75 mL C2H5OH × 500 mL soln = 2.38101mL C H OH 100 mL soln 2 5 Dilute 23.8 mL ethanol with enough water to give a final volume of 500 mL. 4-28. (a) 21.0 g C3H8O3  1000 mL  2.50 L soln = 5.25102 g C H O 100 mL soln 1 L 3 8 3 Weigh 525 g glycerol and add enough water to give a final volume of 2.50 L. 21.0% (w/w) C H O = 21.0 g C3H8O3  1000 g ×2.50 kg soln=5.25 102 g C H O 3 8 3 100 g soln 1 kg 3 8 3 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 18 (b) 2.50 kg soln=0.525 kg C3H8O3 + x kg water x kg water=2.50 kg soln − 0.525 kg C3H8O3 =1.98 kg water Mix 525 g glycerol with 1.98 kg water. Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 19 3 8 3 3 4 (c) 21.0% (v/v) C3H8O3 = 21.0 mL C3H8O3  1000 mL 100 mL soln 1 L × 2.50 L soln = 5.25102 mL C H O Dilute 525 mL glycerol with enough water to give a final volume of 2.50 L. 4-29. 6.00 mol H3PO 4 × 1 L ×750 mL = 4.50 mol H PO L 1000 mL 3 4 86 g H3PO4 ×1.71 g reagent × g water ×1000 mL × mol H3PO4 100 g reagent g water mL L 98.0 g 1.50×101mol H PO = 3 4 L volume 86% (w/w) H PO required = 4.50 mol H PO × L = 3.00 10−1L 3 4 3 4 1.50×101mol H PO 4-30. 3.00 mol HNO 3 × L ×900 mL=2.70 mol HNO L 1000 mL 3 70.5 g HNO3 ×1.42 g reagent ×1 g water ×1000 mL × mol HNO3 100 g reagent g water 1 mL 1 L 63.0 g 1.59×101mol HNO = 3 L volume 70.5% HNO3 required = 2.70 mol HNO3 × L 1.59×101mol HNO = 1.70 10−1L Dilute 170 mL of the concentrated reagent to 900 mL with water. 0.0750 M AgNO 3 = 0.0750 mol AgNO 3 L 4-31. (a) = 0.0750 mol AgNO3 ×169.87 g AgNO3 × 1 L ×500 mL 3 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 20 L 1 mol 1000 mL = 6.37 g AgNO 3 Dissolve 6.37 g AgNO 3 in enough water to give a final volume of 500 mL. Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 21 (b) 0.285 mol HCl ×1 L = 0.285 mol HCl L 0.285 mol HCl × 1 L 6.00 mol HCl = 4.75×10−2 L HCl Take 47.5 mL of the 6.00 M HCl and dilute to 1.00 L with water. 0.0810 mol K+ 1 L −2 + (c) × ×400 mL = 3.24 10 mol K L 1000 mL 3.24 10−2 mol K+×1 mol K4Fe(CN)6 4 mol K+ × 368.43 g K4Fe(CN)6 mol = 2.98 g K4Fe(CN)6 Dissolve 2.98 g K4Fe(CN) 6 in enough water to give a final volume of 400 mL. (d) 3.00 g BaCl2 100 mL soln × 600 mL = 1.8×101g BaCl 1.8101 g BaCl ×1 mol BaCl2 × L = 2.16 10−1 L 2 208.23 g 0.400 mol BaCl Take 216 mL of the 0.400 M BaCl 2 solution and dilute to 600 mL with water (e) 0.120 mol HClO4 ×2.00 L = 0.240 mol HClO L 4 71 g HClO4 ×1.67 g reagent ×1 g water ×1000 mL × mol HClO4 100 g reagent 1 g water 1 mL 1 L 100.46 g 1.18×101mol HClO = 4 L volume 71% (w/w) HClO required = 0.240 mol HClO × L = 2.03×10−2 L 4 4 1.18×101mol HClO Take 20.3 mL of the concentrated reagent and dilute to 2.00 L with water. 2 2 4 Fundamentals of Analytical Chemistry: 9th ed. Chapter 4 22 2 + 60 ppm Na+ = 60 mg Na+ L soln 60 mg Na+ ×9.00 L = 5.4×10 mg Na (f) L soln 2 + 1 g 1 mol Na+ −2 + 5.4 10 mg Na × × = 2.3510 mol Na 1000 mg 22.99 g 2.3510−2 mol Na+ × 1 mol Na2SO4 2 mol Na+ × 142.04 g Na2SO4 1 mol = 1.7 g Na2SO 4 Dissolve 1.7 g Na2SO 4 in enough water to give a final volume of 9.00 L. 4-32. (a) 0.0500 mol KMnO4 × 5.00 L × 158.03 g KMnO4 = 3.95×101g KMnO L 1 mol 4 Dissolve 39.5 g KMnO 4 in enough water to give a final volume of 5.00 L. (b) 0.250 mol HClO 4 L × 4.00 L = 1.00 mol HClO 4 1 L 1.00 mol HClO × = 1.2510−1L reagent 4 8.00 mol reagent Take 125 mL of the 8.00 M reagent and dilute a final of volume of 4.00 L with water. 0.0250 mol I−  1 L ×400 mL = 1.00 × 10−2 mol I− (c) 1.00 L 1000 mL  10−2 mol I− × 1 mol MgI2 2 mol I− × 278.11 g MgI2 1 mol = 1.39 g MgI2 Dissolve 1.39 g MgI 2 in enough water to give a final volume of 400 mL 1.00% (w/v) CuSO 4 = 1.00 g CuSO 4 100 mL soln (d) 1.00 g CuSO4 × 200 mL = 2.00 g CuSO 100 mL soln 4