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#19 key

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  • 1. KEY GENERAL CHEMISTRY-I (1411) S.I. # 19 1. How many grams of solute are present in 50.0 mL of 0.360 M K2Cr2O7? M=(mol/L)  mol = (MxL) (0.360 mol/1L)(0.05L)(294.2 g / mol) = 5.30 g K2Cr2O7 2. If 4.28 g of (NH4)2SO4 is dissolved in enough water to form 300 mL of solution, what is the molarity of the solution? (4.28 g)(1mol/132.2g) = 0.0324 mol / 0.3 L = 0.108 M (NH4)2SO4 3. How many milliliters of 0.240M CuSO4 contain 2.25 g of solute? (2.25 g) (1 mol / 159.6 g) (1L / 0.240 mol)(1000 mL / 1L) = 58.7 mL solution 4. Which will have the highest concentration of potassium ion: 0.20 M KCl, 0.15 M K2CrO4, or 0.080 M K3PO4? KCl  K+ + Cl- ; 0.20 M KCl = 0.20 M K+ K2CrO4  2 K+ + CrO42- ; 0.15 M K2CrO4 = 0.30 M K+ K3PO4  3 K+ + PO43- ; 0.080 M K3PO4 = 0.24 M K+ 0.15 M K2CrO4 has the highest K+ concentration. 5. Which will contain the greater number of moles of potassium ion: 30.0 mL of 0.15M K2CrO4 or 25 mL of 0.080M K3PO4? K2CrO4: (0.30 M K+) (0.030 L) = 0.009 mol K+  greater K3PO4: (0.24 M K+) (0.025 L) = 0.006 mol K+ 6. Indicate the concentration of each ion or molecule present in the following solutions: a. 0.22 M NaOH 0.22 M Na+ and 0.22 M OH- b. 0.16 M CaBr2 0.16 M Ca2+ and 0.32 M Br- c. 0.15 M CH3OH 0.15 M CH3OH is a molecular solute d. a mixture of 40.0 mL of 0.15M KClO3 and 35.0 mL of 0.22M Na2SO4. M2 = M1V1/V2 where V2 is the total solution volume. + K : (0.15 M x 0.040 L) / (0.075 L) = 0.080 M ClO3- : concentration ClO3- = K+ = 0.080 M SO42-: (0.22 M x 0.0350L) / (0.075 L) = 0.1027 M SO42- Na+ : concentration of Na+ = 2xSO42- = 0.21 M Na+
  • 2. KEY 7. What mass of NaOH is needed to precipitate the Cd2+ ions from 25.0 mL of 0.500M Cd(NO3)2 solution? Cd(NO3)2(aq) + 2NaOH(aq)  Cd(OH)2(s) + 2NaNO3(aq) (0.025 L) (0.50mol Cd(NO3)2 / 1L) ( 2 mol NaOH / 1mol Cd(NO3)2)(40gNaOH/1mol) = 1g NaOH. 8. How many milliliters of 0.120M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)2 solution? 2HCl(aq) + Ba(OH)2 (aq)  BaCl2(aq) + 2H2O (l) [(0.101 mol Ba(OH)2 /1 L] (0.05 L Ba(OH)2 ) (2 mol HCl / 1 mol Ba(OH)2) x (1L HCl / 0.120mol HCl) = 0.0842 L HCl soln. 9. How many milliliters of 0.125 M H2SO4 are needed to neutralize 0.200 g of NaOH? H2SO4 (aq) + 2NaOH (aq)  Na2SO4 (aq) + 2H2O (l) (0.200 g NaOH) (1mol NaOH / 40.0 g ) (1mol H2SO4 / 2 mol NaOH) x (1L H2SO4/0.125 mol H2SO4) = 0.0200 L or 20 mL H2SO4 soln. 10. If 55.8 mL of BaCl2 solution is needed to precipitate all of the sulfate ion in a 752-mg sample of Na2SO4, what is the molarity of the solution? BaCl2 (aq) + Na2SO4 (aq)  BaSO4 (s) + 2NaCl (aq) 752 mg = 0.752 g Na2SO4 (0.752 g Na2SO4) (1 mol Na2SO4 / 142.1 g )(1mol BaCl2 / 1 mol Na2SO4) (1 / 0.0558L) = 0.0948 M BaCl2

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