{"id":1289,"date":"2021-02-19T14:50:30","date_gmt":"2021-02-19T14:50:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/?post_type=chapter&p=1289"},"modified":"2021-02-26T20:35:20","modified_gmt":"2021-02-26T20:35:20","slug":"exercises-5","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/chapter\/exercises-5\/","title":{"raw":"Exercises","rendered":"Exercises"},"content":{"raw":"

Part 1<\/h3>\r\n
    \r\n \t
  1. Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.<\/li>\r\n \t
  2. What information do we need to calculate the molarity of a sulfuric acid solution?<\/li>\r\n \t
  3. What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different<\/li>\r\n \t
  4. Determine the molarity of each of the following solutions:\r\n
      \r\n \t
    1. 1.457 mol KCl in 1.500 L of solution<\/li>\r\n \t
    2. 0.515 g of H2<\/sub>SO4<\/sub> in 1.00 L of solution<\/li>\r\n \t
    3. 20.54 g of Al(NO3<\/sub>)3<\/sub> in 1575 mL of solution<\/li>\r\n \t
    4. 2.76 kg of CuSO4<\/sub>\u20225H2<\/sub>O in 1.45 L of solution<\/li>\r\n \t
    5. 0.005653 mol of Br2<\/sub> in 10.00 mL of solution<\/li>\r\n \t
    6. 0.000889 g of glycine, C2<\/sub>H5<\/sub>NO2<\/sub>, in 1.05 mL of solution<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    7. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M<\/em> glucose, C6<\/sub>H12<\/sub>O6<\/sub>, used for intravenous injection?\r\n
        \r\n \t
      1. Outline the steps necessary to answer the question.<\/li>\r\n \t
      2. Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      3. Consider this question: What is the mass of solute in 200.0 L of a 1.556-M<\/em> solution of KBr?\r\n
          \r\n \t
        1. Outline the steps necessary to answer the question.<\/li>\r\n \t
        2. Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        3. Calculate the number of moles and the mass of the solute in each of the following solutions:\r\n
            \r\n \t
          1. 2.00 L of 18.5 M<\/em> H2<\/sub>SO4<\/sub>, concentrated sulfuric acid<\/li>\r\n \t
          2. 100.0 mL of [latex]3.8\\times {10}^{-5}M\\text{NaCN,}[\/latex] the minimum lethal concentration of sodium cyanide in blood serum<\/li>\r\n \t
          3. 5.50 L of 13.3 M<\/em> H2<\/sub>CO, the formaldehyde used to \u201cfix\u201d tissue samples<\/li>\r\n \t
          4. 325 mL of [latex]1.8\\times {10}^{-6}M[\/latex] FeSO4<\/sub>, the minimum concentration of iron sulfate detectable by taste in drinking water<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
          5. Calculate the number of moles and the mass of the solute in each of the following solutions:\r\n
              \r\n \t
            1. 325 mL of [latex]8.23\\times {10}^{-5}M\\text{KI}[\/latex], a source of iodine in the diet<\/li>\r\n \t
            2. 75.0 mL of [latex]2.2\\times {10}^{-5}M{\\text{H}}_{2}{\\text{SO}}_{4}[\/latex], a sample of acid rain<\/li>\r\n \t
            3. 0.2500 L of 0.1135 M<\/em> K2<\/sub>CrO4<\/sub>, an analytical reagent used in iron assays<\/li>\r\n \t
            4. 10.5 L of 3.716 M<\/em> (NH4<\/sub>)2<\/sub>SO4<\/sub>, a liquid fertilizer<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            5. Consider this question: What is the molarity of KMnO4<\/sub> in a solution of 0.0908 g of KMnO4<\/sub> in 0.500 L of solution?\r\n
                \r\n \t
              1. Outline the steps necessary to answer the question.<\/li>\r\n \t
              2. Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              3. Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?\r\n
                  \r\n \t
                1. Outline the steps necessary to answer the question.<\/li>\r\n \t
                2. Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                3. Calculate the molarity of each of the following solutions:\r\n
                    \r\n \t
                  1. 0.195 g of cholesterol, C27<\/sub>H46<\/sub>O, in 0.100 L of serum, the average concentration of cholesterol in human serum<\/li>\r\n \t
                  2. 4.25 g of NH3<\/sub> in 0.500 L of solution, the concentration of NH3<\/sub> in household ammonia<\/li>\r\n \t
                  3. 1.49 kg of isopropyl alcohol, C3<\/sub>H7<\/sub>OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol<\/li>\r\n \t
                  4. 0.029 g of I2<\/sub> in 0.100 L of solution, the solubility of I2<\/sub> in water at 20 \u00b0C<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                  5. Calculate the molarity of each of the following solutions:\r\n
                      \r\n \t
                    1. 293 g HCl in 666 mL of solution, a concentrated HCl solution<\/li>\r\n \t
                    2. 2.026 g FeCl3<\/sub> in 0.1250 L of a solution used as an unknown in general chemistry laboratories<\/li>\r\n \t
                    3. 0.001 mg Cd2+<\/sup> in 0.100 L, the maximum permissible concentration of cadmium in drinking water<\/li>\r\n \t
                    4. 0.0079 g C7<\/sub>H5<\/sub>SNO3<\/sub> in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                    5. There is about 1.0 g of calcium, as Ca2+<\/sup>, in 1.0 L of milk. What is the molarity of Ca2+<\/sup> in milk?<\/li>\r\n \t
                    6. What volume of a 1.00-M<\/em> Fe(NO3<\/sub>)3<\/sub> solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M<\/em>?<\/li>\r\n \t
                    7. If 0.1718 L of a 0.3556-M<\/em> C3<\/sub>H7<\/sub>OH solution is diluted to a concentration of 0.1222 M<\/em>, what is the volume of the resulting solution?<\/li>\r\n \t
                    8. If 4.12 L of a 0.850 M<\/em>-H3<\/sub>PO4<\/sub> solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?<\/li>\r\n \t
                    9. What volume of a 0.33-M<\/em> C12<\/sub>H22<\/sub>O11<\/sub> solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M<\/em>?<\/li>\r\n \t
                    10. What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M<\/em> solution is allowed to evaporate until the volume is reduced to 0.105 L?<\/li>\r\n \t
                    11. A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?<\/li>\r\n \t
                    12. An experiment in a general chemistry laboratory calls for a 2.00-M<\/em> solution of HCl. How many mL of 11.9 M<\/em> HCl would be required to make 250 mL of 2.00 M<\/em> HCl?<\/li>\r\n \t
                    13. What volume of a 0.20-M<\/em> K2<\/sub>SO4<\/sub> solution contains 57 g of K2<\/sub>SO4<\/sub>?<\/li>\r\n \t
                    14. The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg\/L. If an industry is discharging hexavalent chromium as potassium dichromate (K2<\/sub>Cr2<\/sub>O7<\/sub>), what is the maximum permissible molarity of that substance?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"812628\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"812628\"]\r\n\r\n2.\u00a0We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.\r\n\r\n6. The answers are as follows:\r\n
                        \r\n \t
                      1. determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass;<\/li>\r\n \t
                      2. 0.500 L contains [latex]0.30M\\times 0.500\\text{L}=1.5\\times {10}^{-1}\\text{mol.}[\/latex] Molar mass (glucose): [latex]6\\times \\text{12.0011 g}+12\\times \\text{1.00794 g}+6\\times \\text{15.9994 g}=\\text{180.158 g},1.5\\times {10}^{-1}\\cancel{\\text{mol}}\\times \\text{180.158 g\/}\\cancel{\\text{mol}}=\\text{27 g.}[\/latex]<\/li>\r\n<\/ol>\r\n8. The molarity must be converted to moles of solute, which is then converted to grams of solute:\r\n\"Three\r\n

                        [latex]M=\\frac{\\text{mol}}{\\text{liter}}\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\text{mol}=M\\times \\text{liter}[\/latex]<\/p>\r\n\r\n

                          \r\n \t
                        1. [latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}{\\text{SO}}_{4}=2.00\\cancel{\\text{L}}\\times \\frac{18.5\\text{mol}}{\\cancel{\\text{L}}}=37.0\\text{mol}{\\text{H}}_{2}{\\text{SO}}_{4}\\\\ 37.0\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}\\times \\frac{98.08{\\text{g H}}_{2}{\\text{SO}}_{4}}{1\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}}=3.63\\times {10}^{3}{\\text{g H}}_{2}{\\text{SO}}_{4}\\end{array}[\/latex]<\/li>\r\n \t
                        2. [latex]\\begin{array}{l}\\\\ \\text{mol NaCN}=0.1000\\cancel{\\text{L}}\\times \\frac{3.8\\times {10}^{-5}\\text{mol}}{\\cancel{\\text{L}}}=3.8\\times {10}^{-6}\\text{mol NaCN}\\\\ 3.8\\times {10}^{-5}\\cancel{\\text{mol NaCN}}\\times \\frac{49.01\\text{g}}{1\\cancel{\\text{mol NaCN}}}=1.9\\times {\\text{10}}^{-4}\\text{g NaCN}\\end{array}[\/latex]<\/li>\r\n \t
                        3. [latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}\\text{CO}=5.50\\cancel{\\text{L}}\\times \\frac{13.3\\text{mol}}{\\cancel{\\text{L}}}=73.2\\text{mol}{\\text{H}}_{2}\\text{CO}\\\\ 73.2\\cancel{\\text{mol}{\\text{H}}_{2}\\text{CO}}\\times \\frac{30.026\\text{g}}{1\\cancel{{\\text{mol H}}_{2}\\text{CO}}}=2198\\text{g}{\\text{H}}_{2}\\text{CO}=2.20\\text{kg}{\\text{H}}_{2}\\text{CO}\\end{array}[\/latex]<\/li>\r\n \t
                        4. [latex]\\begin{array}{l}\\\\ {\\text{mol FeSO}}_{4}=0.325\\cancel{\\text{L}}\\times \\frac{1.8\\times {10}^{-6}\\text{mol}}{\\cancel{L}}=5.9\\times {10}^{-7}{\\text{mol FeSO}}_{4}\\\\ 5.85\\times {10}^{-7}\\cancel{{\\text{mol FeSO}}_{4}}\\times \\frac{151.9\\text{g}}{1\\cancel{{\\text{mol FeSO}}_{4}}}=8.9\\times {10}^{-5}{\\text{g FeSO}}_{4}\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n10. The answers are as follows:\r\n
                            \r\n \t
                          1. Determine the molar mass of KMnO4<\/sub>; determine the number of moles of KMnO4<\/sub> in the solution; from the number of moles and the volume of solution, determine the molarity<\/li>\r\n \t
                          2. Molar mass of KMnO4<\/sub> = 158.0264 g\/mol\r\n[latex]\\begin{array}{l}\\\\ \\text{mol}{\\text{KMnO}}_{4}=0.0908\\cancel{\\text{g}{\\text{KMnO}}_{4}}\\times \\frac{\\text{1 mol}}{158.0264\\cancel{{\\text{g KMnO}}_{4}}}=5.746\\times {10}^{-4}\\text{mol}\\\\ M{\\text{KMnO}}_{4}=\\frac{5.746\\times {10}^{-4}\\text{mol}}{0.500\\text{L}}=1.15\\times {10}^{-3}M\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n12.\u00a0The molarity for each solution is as follows:\r\n
                              \r\n \t
                            1. [latex]M{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.195\\cancel{g}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}{386.660\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}}{0.100\\text{L}}=5.04\\times {10}^{-3}M[\/latex]<\/li>\r\n \t
                            2. [latex]M{\\text{NH}}_{3}=\\frac{\\text{mol}}{V}=\\frac{\\frac{4.25\\cancel{g}{\\text{NH}}_{3}}{17.0304\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{NH}}_{3}}}{0.500\\text{L}}=0.499M[\/latex]<\/li>\r\n \t
                            3. [latex]M{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}=\\frac{\\text{mol}}{V}=\\frac{1.49\\cancel{\\text{kg}}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}\\times \\frac{1000\\cancel{g}}{1\\cancel{\\text{kg}}}\\times \\frac{1\\text{mol}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}}{60.096\\cancel{\\text{g}}}}{2.50\\text{L}}=9.92M[\/latex]<\/li>\r\n \t
                            4. [latex]M{\\text{I}}_{2}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.029\\cancel{\\text{g}}{\\text{I}}_{2}}{253.8090\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{I}}_{2}}}{0.100\\text{L}}=1.1\\times {10}^{-3}M[\/latex]<\/li>\r\n<\/ol>\r\n14.\u00a0[latex]M=\\frac{\\text{mol}}{V}=\\frac{\\frac{1.0\\cancel{\\text{g}}}{40.08\\cancel{\\text{g}}{\\text{mol}}^{-1}}}{1.0\\text{L}}=0.025M[\/latex]\r\n\r\n16.\u00a0[latex]\\begin{array}{c}\\frac{{C}_{1}{V}_{1}}{{C}_{2}}={V}_{2}\\\\ \\frac{\\frac{0.3556\\text{mol}}{\\text{L}}\\times 0.1718\\text{L}}{\\frac{0.1222\\text{mol}}{\\text{L}}}={V}_{2}\\\\ 0.5000\\text{L}={V}_{2}\\end{array}[\/latex]\r\n\r\n18.\u00a0[latex]{V}_{1}=\\frac{{V}_{2}\\times {M}_{2}}{{M}_{2}}=25\\text{mL}\\times \\frac{0.025M}{0.33M}=1.9\\text{mL}[\/latex]\r\n\r\n22. Determine the number of moles in 434.4 g of HCl: 1.00794 + 35.4527 = 36.4606 g\/mol\r\n

                              [latex]\\text{mol HCl}=\\frac{434.4\\cancel{g}}{36.4606\\cancel{g}{\\text{mol}}^{-1}}=11.91\\text{mol}[\/latex]<\/p>\r\nThis HCl is present in 1.00 L, so the molarity is 11.9 M<\/em>.\r\n\r\n24.\u00a0[latex]57\\text{g}{\\text{K}}_{2}{\\text{SO}}_{4}\\times \\frac{1\\text{mol}}{174.26\\text{g}}\\times \\frac{1\\text{L}}{0.20\\text{mol}}=1.6\\text{L}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nPart 2\r\n

                                \r\n \t
                              1. Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO3<\/sub> by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3<\/sub> by mass?\r\n
                                  \r\n \t
                                1. Outline the steps necessary to answer the question.<\/li>\r\n \t
                                2. Answer the question.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                3. What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?<\/li>\r\n \t
                                4. What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g\/mL.<\/li>\r\n \t
                                5. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm\u20133<\/sup> and contains 37.21% HCl by mass?<\/li>\r\n \t
                                6. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of CaCO3<\/sub>, which is equivalent to milligrams of CaCO3<\/sub> per liter of water. What is the molar concentration of Ca2+ <\/sup>ions in a water sample with a hardness count of 175 mg CaCO3<\/sub>\/L?<\/li>\r\n \t
                                7. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g\/mL and calculate the molarity of mercury in the stream.<\/li>\r\n \t
                                8. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM<\/em> is observed, what is the concentration of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) in mg\/dL?<\/li>\r\n \t
                                9. A cough syrup contains 5.0% ethyl alcohol, C2<\/sub>H5<\/sub>OH, by mass. If the density of the solution is 0.9928 g\/mL, determine the molarity of the alcohol in the cough syrup.<\/li>\r\n \t
                                10. D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose (C6<\/sub>H12<\/sub>O6<\/sub>) in water. If the density of D5W is 1.029 g\/mL, calculate the molarity of dextrose in the solution.<\/li>\r\n \t
                                11. Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H2<\/sub>SO4<\/sub>, for which the density is 1.3057 g\/mL.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"361643\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"361643\"]\r\n\r\n1.\u00a0The answers are as follows:\r\n
                                    \r\n \t
                                  1. The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:\r\n[latex]\\%{\\text{mass}}_{1}\\times {\\text{mass}}_{1}=\\%{\\text{mass}}_{2}\\times {\\text{mass}}_{2}[\/latex]\r\nThis equation can be rearranged to isolate mass1<\/sub> and the given quantities substituted into this equation.<\/li>\r\n \t
                                  2. [latex]{\\text{mass}}_{1}=\\frac{\\%{\\text{mass}}_{2}\\times {\\text{mass}}_{2}}{\\%{\\text{mass}}_{1}}=\\frac{10.0\\%\\times 400.0\\text{g}}{68.0\\%}=58.8\\text{g}[\/latex]<\/li>\r\n<\/ol>\r\n3. The mass of the 10% solution is\u00a0[latex]1000{\\cancel{\\text{cm}}}^{3}\\times \\frac{1.109\\text{g}}{{\\cancel{\\text{cm}}}^{3}}=1.11\\times {10}^{3}\\text{g.}[\/latex]\r\n\r\nThe mass of pure NaOH required is\u00a0[latex]\\text{mass}\\left(\\text{NaOH}\\right)=\\frac{10.0\\%}{100.0\\%}\\times 1.11\\times {10}^{3}\\text{g}=1.11\\times {10}^{2}\\text{g.}[\/latex]\r\n\r\nThis mass of NaOH must come from the 97.0% solution:\u00a0[latex]\\begin{array}{l}\\\\ \\text{mass}\\left(\\text{NaOH solution}\\right)=\\frac{97.0\\%}{100.0\\%}=1.11\\times {10}^{2}\\text{g}\\\\ \\text{mass}\\left(\\text{NaOH solution}\\right)=\\frac{1.11\\times {10}^{2}\\text{g}}{0.970}=114\\text{g}\\end{array}[\/latex]\r\n\r\n5. \u00a0Since CaCO3<\/sub> contains 1 mol Ca2+<\/sup> per mol of CaCO3<\/sub>, the molar concentration of Ca2+<\/sup> equals the molarity of CaCO3<\/sub>\r\n

                                    [latex]M{\\text{Ca}}^{2+}=\\frac{\\text{mol}{\\text{CaCO}}_{3}}{\\text{L}}=\\frac{175\\cancel{\\text{mg}}\\times \\left(\\frac{1\\text{mol}}{100.0792\\cancel{\\text{g}}}\\right)\\times \\left(\\frac{1\\cancel{\\text{g}}}{1000\\text{mg}}\\right)}{1\\text{L}}=1.75\\times {10}^{-3}M[\/latex]<\/p>\r\n7. 1 mg\/dL = 0.01 g\/L and 1 L = 10 dL\r\n

                                    [latex]5.3\\cancel{\\text{mmol}}\\text{\/L}\\times 180.158\\text{mg\/}\\cancel{\\text{mmol}}=9.5\\times {10}^{2}\\text{mg\/L}[\/latex]\r\n[latex]9.5\\times {10}^{2}\\text{mg\/L}\\times \\frac{1\\text{L}}{10\\text{dL}}=95\\text{mg\/dL}[\/latex]<\/p>\r\n11. The molar mass of C6<\/sub>H12<\/sub>O6<\/sub> is [latex]6\\times 12.011+12\\times 1.00794+6\\times 15.9994=180.2\\text{g\/mol.}[\/latex] In 1.000 L, there are:\r\n

                                    [latex]\\begin{array}{l}\\left(1000\\cancel{\\text{mL}}\\times 1.029\\text{g}{\\cancel{\\text{mL}}}^{-1}\\right)=1029\\text{g}\\\\ \\text{mol dextrose}=1029\\cancel{\\text{g}}\\times 0.050\\times \\frac{1\\text{mol}}{180.2\\cancel{\\text{g}}}=0.29\\text{mol}\\end{array}[\/latex] C6<\/sub>H12<\/sub>O6<\/sub>.<\/p>\r\nSince we selected the volume to be 1.00 L, the molarity of dextrose is\u00a0[latex]\\text{molarity}=\\frac{\\text{mol}}{\\text{L}}=\\frac{0.29\\text{mol}}{1.00\\text{L}}=0.29\\text{mol}[\/latex].\r\n\r\n[\/hidden-answer]\r\n

                                    Part 3<\/h3>\r\n
                                      \r\n \t
                                    1. Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.<\/li>\r\n \t
                                    2. Consider the solutions presented:\r\n\"In\r\n
                                        \r\n \t
                                      1. Which of the following sketches best represents the ions in a solution of Fe(NO3<\/sub>)3<\/sub>(aq<\/em>)?<\/li>\r\n \t
                                      2. Write a balanced chemical equation showing the products of the dissolution of Fe(NO3<\/sub>)3<\/sub>.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                      3. Compare the processes that occur when methanol (CH3<\/sub>OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.<\/li>\r\n \t
                                      4. What is the expected electrical conductivity of the following solutions?\r\n
                                          \r\n \t
                                        1. NaOH(aq<\/em>)<\/li>\r\n \t
                                        2. HCl(aq<\/em>)<\/li>\r\n \t
                                        3. C6<\/sub>H12<\/sub>O6<\/sub>(aq<\/em>) (glucose)<\/li>\r\n \t
                                        4. NH3<\/sub>(l<\/em>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                        5. Why are most solid<\/em> ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid<\/em> (molten) ionic compound to be electrically conductive or nonconductive? Explain.<\/li>\r\n \t
                                        6. Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:\r\n
                                            \r\n \t
                                          1. the solutions in Figure\u00a02<\/li>\r\n \t
                                          2. methanol, CH3<\/sub>OH, dissolved in ethanol, C2<\/sub>H5<\/sub>OH<\/li>\r\n \t
                                          3. methane, CH4<\/sub>, dissolved in benzene, C6<\/sub>H6\u00a0<\/sub><\/li>\r\n \t
                                          4. the polar halocarbon CF2<\/sub>Cl2<\/sub> dissolved in the polar halocarbon CF2<\/sub>ClCFCl2\u00a0<\/sub><\/li>\r\n \t
                                          5. O2<\/sub>(l<\/em>) in N2<\/sub>(l<\/em>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"43612\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"43612\"]\r\n\r\n3. The answers are as follows:\r\n
                                              \r\n \t
                                            1. Fe(NO3<\/sub>)3<\/sub> is a strong electrolyte, thus it should completely dissociate into Fe3+<\/sup> and (NO3<\/sub>\u2212<\/sup>) ions. Therefore, (z<\/em>) best represents the solution.<\/li>\r\n \t
                                            2. Fe(NO3<\/sub>)3<\/sub>(s<\/em>) \u27f6 Fe3+<\/sup>(aq<\/em>) + 3NO3<\/sub>\u2212<\/sup>(aq<\/em>)<\/li>\r\n<\/ol>\r\n5. The expected electrical conductivity is as follows:\r\n
                                                \r\n \t
                                              1. high conductivity (solute is an ionic compound that will dissociate when dissolved);<\/li>\r\n \t
                                              2. high conductivity (solute is a strong acid and will ionize completely when dissolved);<\/li>\r\n \t
                                              3. nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water);<\/li>\r\n \t
                                              4. low conductivity (solute is a weak base and will partially ionize when dissolved)<\/li>\r\n<\/ol>\r\n7. The types of intermolecular attraction are as follows:\r\n
                                                  \r\n \t
                                                1. ion-dipole;<\/li>\r\n \t
                                                2. hydrogen bonds;<\/li>\r\n \t
                                                3. dispersion forces;<\/li>\r\n \t
                                                4. dipole-dipole attractions;<\/li>\r\n \t
                                                5. dispersion forces<\/li>\r\n<\/ol>\r\n[\/hidden-answer]","rendered":"

                                                  Part 1<\/h3>\n
                                                    \n
                                                  1. Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.<\/li>\n
                                                  2. What information do we need to calculate the molarity of a sulfuric acid solution?<\/li>\n
                                                  3. What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different<\/li>\n
                                                  4. Determine the molarity of each of the following solutions:\n
                                                      \n
                                                    1. 1.457 mol KCl in 1.500 L of solution<\/li>\n
                                                    2. 0.515 g of H2<\/sub>SO4<\/sub> in 1.00 L of solution<\/li>\n
                                                    3. 20.54 g of Al(NO3<\/sub>)3<\/sub> in 1575 mL of solution<\/li>\n
                                                    4. 2.76 kg of CuSO4<\/sub>\u20225H2<\/sub>O in 1.45 L of solution<\/li>\n
                                                    5. 0.005653 mol of Br2<\/sub> in 10.00 mL of solution<\/li>\n
                                                    6. 0.000889 g of glycine, C2<\/sub>H5<\/sub>NO2<\/sub>, in 1.05 mL of solution<\/li>\n<\/ol>\n<\/li>\n
                                                    7. Consider this question: What is the mass of the solute in 0.500 L of 0.30 M<\/em> glucose, C6<\/sub>H12<\/sub>O6<\/sub>, used for intravenous injection?\n
                                                        \n
                                                      1. Outline the steps necessary to answer the question.<\/li>\n
                                                      2. Answer the question.<\/li>\n<\/ol>\n<\/li>\n
                                                      3. Consider this question: What is the mass of solute in 200.0 L of a 1.556-M<\/em> solution of KBr?\n
                                                          \n
                                                        1. Outline the steps necessary to answer the question.<\/li>\n
                                                        2. Answer the question.<\/li>\n<\/ol>\n<\/li>\n
                                                        3. Calculate the number of moles and the mass of the solute in each of the following solutions:\n
                                                            \n
                                                          1. 2.00 L of 18.5 M<\/em> H2<\/sub>SO4<\/sub>, concentrated sulfuric acid<\/li>\n
                                                          2. 100.0 mL of [latex]3.8\\times {10}^{-5}M\\text{NaCN,}[\/latex] the minimum lethal concentration of sodium cyanide in blood serum<\/li>\n
                                                          3. 5.50 L of 13.3 M<\/em> H2<\/sub>CO, the formaldehyde used to \u201cfix\u201d tissue samples<\/li>\n
                                                          4. 325 mL of [latex]1.8\\times {10}^{-6}M[\/latex] FeSO4<\/sub>, the minimum concentration of iron sulfate detectable by taste in drinking water<\/li>\n<\/ol>\n<\/li>\n
                                                          5. Calculate the number of moles and the mass of the solute in each of the following solutions:\n
                                                              \n
                                                            1. 325 mL of [latex]8.23\\times {10}^{-5}M\\text{KI}[\/latex], a source of iodine in the diet<\/li>\n
                                                            2. 75.0 mL of [latex]2.2\\times {10}^{-5}M{\\text{H}}_{2}{\\text{SO}}_{4}[\/latex], a sample of acid rain<\/li>\n
                                                            3. 0.2500 L of 0.1135 M<\/em> K2<\/sub>CrO4<\/sub>, an analytical reagent used in iron assays<\/li>\n
                                                            4. 10.5 L of 3.716 M<\/em> (NH4<\/sub>)2<\/sub>SO4<\/sub>, a liquid fertilizer<\/li>\n<\/ol>\n<\/li>\n
                                                            5. Consider this question: What is the molarity of KMnO4<\/sub> in a solution of 0.0908 g of KMnO4<\/sub> in 0.500 L of solution?\n
                                                                \n
                                                              1. Outline the steps necessary to answer the question.<\/li>\n
                                                              2. Answer the question.<\/li>\n<\/ol>\n<\/li>\n
                                                              3. Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?\n
                                                                  \n
                                                                1. Outline the steps necessary to answer the question.<\/li>\n
                                                                2. Answer the question.<\/li>\n<\/ol>\n<\/li>\n
                                                                3. Calculate the molarity of each of the following solutions:\n
                                                                    \n
                                                                  1. 0.195 g of cholesterol, C27<\/sub>H46<\/sub>O, in 0.100 L of serum, the average concentration of cholesterol in human serum<\/li>\n
                                                                  2. 4.25 g of NH3<\/sub> in 0.500 L of solution, the concentration of NH3<\/sub> in household ammonia<\/li>\n
                                                                  3. 1.49 kg of isopropyl alcohol, C3<\/sub>H7<\/sub>OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol<\/li>\n
                                                                  4. 0.029 g of I2<\/sub> in 0.100 L of solution, the solubility of I2<\/sub> in water at 20 \u00b0C<\/li>\n<\/ol>\n<\/li>\n
                                                                  5. Calculate the molarity of each of the following solutions:\n
                                                                      \n
                                                                    1. 293 g HCl in 666 mL of solution, a concentrated HCl solution<\/li>\n
                                                                    2. 2.026 g FeCl3<\/sub> in 0.1250 L of a solution used as an unknown in general chemistry laboratories<\/li>\n
                                                                    3. 0.001 mg Cd2+<\/sup> in 0.100 L, the maximum permissible concentration of cadmium in drinking water<\/li>\n
                                                                    4. 0.0079 g C7<\/sub>H5<\/sub>SNO3<\/sub> in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.<\/li>\n<\/ol>\n<\/li>\n
                                                                    5. There is about 1.0 g of calcium, as Ca2+<\/sup>, in 1.0 L of milk. What is the molarity of Ca2+<\/sup> in milk?<\/li>\n
                                                                    6. What volume of a 1.00-M<\/em> Fe(NO3<\/sub>)3<\/sub> solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M<\/em>?<\/li>\n
                                                                    7. If 0.1718 L of a 0.3556-M<\/em> C3<\/sub>H7<\/sub>OH solution is diluted to a concentration of 0.1222 M<\/em>, what is the volume of the resulting solution?<\/li>\n
                                                                    8. If 4.12 L of a 0.850 M<\/em>-H3<\/sub>PO4<\/sub> solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?<\/li>\n
                                                                    9. What volume of a 0.33-M<\/em> C12<\/sub>H22<\/sub>O11<\/sub> solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M<\/em>?<\/li>\n
                                                                    10. What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M<\/em> solution is allowed to evaporate until the volume is reduced to 0.105 L?<\/li>\n
                                                                    11. A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?<\/li>\n
                                                                    12. An experiment in a general chemistry laboratory calls for a 2.00-M<\/em> solution of HCl. How many mL of 11.9 M<\/em> HCl would be required to make 250 mL of 2.00 M<\/em> HCl?<\/li>\n
                                                                    13. What volume of a 0.20-M<\/em> K2<\/sub>SO4<\/sub> solution contains 57 g of K2<\/sub>SO4<\/sub>?<\/li>\n
                                                                    14. The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg\/L. If an industry is discharging hexavalent chromium as potassium dichromate (K2<\/sub>Cr2<\/sub>O7<\/sub>), what is the maximum permissible molarity of that substance?<\/li>\n<\/ol>\n
                                                                      Show Selected Answers<\/span><\/p>\n
                                                                      \n

                                                                      2.\u00a0We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.<\/p>\n

                                                                      6. The answers are as follows:<\/p>\n

                                                                        \n
                                                                      1. determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass;<\/li>\n
                                                                      2. 0.500 L contains [latex]0.30M\\times 0.500\\text{L}=1.5\\times {10}^{-1}\\text{mol.}[\/latex] Molar mass (glucose): [latex]6\\times \\text{12.0011 g}+12\\times \\text{1.00794 g}+6\\times \\text{15.9994 g}=\\text{180.158 g},1.5\\times {10}^{-1}\\cancel{\\text{mol}}\\times \\text{180.158 g\/}\\cancel{\\text{mol}}=\\text{27 g.}[\/latex]<\/li>\n<\/ol>\n

                                                                        8. The molarity must be converted to moles of solute, which is then converted to grams of solute:
                                                                        \n\"Three<\/p>\n

                                                                        [latex]M=\\frac{\\text{mol}}{\\text{liter}}\\,\\,\\,\\,\\text{or}\\,\\,\\,\\,\\text{mol}=M\\times \\text{liter}[\/latex]<\/p>\n

                                                                          \n
                                                                        1. [latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}{\\text{SO}}_{4}=2.00\\cancel{\\text{L}}\\times \\frac{18.5\\text{mol}}{\\cancel{\\text{L}}}=37.0\\text{mol}{\\text{H}}_{2}{\\text{SO}}_{4}\\\\ 37.0\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}\\times \\frac{98.08{\\text{g H}}_{2}{\\text{SO}}_{4}}{1\\cancel{{\\text{mol H}}_{2}{\\text{SO}}_{4}}}=3.63\\times {10}^{3}{\\text{g H}}_{2}{\\text{SO}}_{4}\\end{array}[\/latex]<\/li>\n
                                                                        2. [latex]\\begin{array}{l}\\\\ \\text{mol NaCN}=0.1000\\cancel{\\text{L}}\\times \\frac{3.8\\times {10}^{-5}\\text{mol}}{\\cancel{\\text{L}}}=3.8\\times {10}^{-6}\\text{mol NaCN}\\\\ 3.8\\times {10}^{-5}\\cancel{\\text{mol NaCN}}\\times \\frac{49.01\\text{g}}{1\\cancel{\\text{mol NaCN}}}=1.9\\times {\\text{10}}^{-4}\\text{g NaCN}\\end{array}[\/latex]<\/li>\n
                                                                        3. [latex]\\begin{array}{l}\\\\ {\\text{mol H}}_{2}\\text{CO}=5.50\\cancel{\\text{L}}\\times \\frac{13.3\\text{mol}}{\\cancel{\\text{L}}}=73.2\\text{mol}{\\text{H}}_{2}\\text{CO}\\\\ 73.2\\cancel{\\text{mol}{\\text{H}}_{2}\\text{CO}}\\times \\frac{30.026\\text{g}}{1\\cancel{{\\text{mol H}}_{2}\\text{CO}}}=2198\\text{g}{\\text{H}}_{2}\\text{CO}=2.20\\text{kg}{\\text{H}}_{2}\\text{CO}\\end{array}[\/latex]<\/li>\n
                                                                        4. [latex]\\begin{array}{l}\\\\ {\\text{mol FeSO}}_{4}=0.325\\cancel{\\text{L}}\\times \\frac{1.8\\times {10}^{-6}\\text{mol}}{\\cancel{L}}=5.9\\times {10}^{-7}{\\text{mol FeSO}}_{4}\\\\ 5.85\\times {10}^{-7}\\cancel{{\\text{mol FeSO}}_{4}}\\times \\frac{151.9\\text{g}}{1\\cancel{{\\text{mol FeSO}}_{4}}}=8.9\\times {10}^{-5}{\\text{g FeSO}}_{4}\\end{array}[\/latex]<\/li>\n<\/ol>\n

                                                                          10. The answers are as follows:<\/p>\n

                                                                            \n
                                                                          1. Determine the molar mass of KMnO4<\/sub>; determine the number of moles of KMnO4<\/sub> in the solution; from the number of moles and the volume of solution, determine the molarity<\/li>\n
                                                                          2. Molar mass of KMnO4<\/sub> = 158.0264 g\/mol
                                                                            \n[latex]\\begin{array}{l}\\\\ \\text{mol}{\\text{KMnO}}_{4}=0.0908\\cancel{\\text{g}{\\text{KMnO}}_{4}}\\times \\frac{\\text{1 mol}}{158.0264\\cancel{{\\text{g KMnO}}_{4}}}=5.746\\times {10}^{-4}\\text{mol}\\\\ M{\\text{KMnO}}_{4}=\\frac{5.746\\times {10}^{-4}\\text{mol}}{0.500\\text{L}}=1.15\\times {10}^{-3}M\\end{array}[\/latex]<\/li>\n<\/ol>\n

                                                                            12.\u00a0The molarity for each solution is as follows:<\/p>\n

                                                                              \n
                                                                            1. [latex]M{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.195\\cancel{g}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}{386.660\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{C}}_{27}{\\text{H}}_{46}\\text{O}}}{0.100\\text{L}}=5.04\\times {10}^{-3}M[\/latex]<\/li>\n
                                                                            2. [latex]M{\\text{NH}}_{3}=\\frac{\\text{mol}}{V}=\\frac{\\frac{4.25\\cancel{g}{\\text{NH}}_{3}}{17.0304\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{NH}}_{3}}}{0.500\\text{L}}=0.499M[\/latex]<\/li>\n
                                                                            3. [latex]M{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}=\\frac{\\text{mol}}{V}=\\frac{1.49\\cancel{\\text{kg}}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}\\times \\frac{1000\\cancel{g}}{1\\cancel{\\text{kg}}}\\times \\frac{1\\text{mol}{\\text{C}}_{3}{\\text{H}}_{7}\\text{OH}}{60.096\\cancel{\\text{g}}}}{2.50\\text{L}}=9.92M[\/latex]<\/li>\n
                                                                            4. [latex]M{\\text{I}}_{2}=\\frac{\\text{mol}}{V}=\\frac{\\frac{0.029\\cancel{\\text{g}}{\\text{I}}_{2}}{253.8090\\cancel{\\text{g}}{\\text{mol}}^{-1}{\\text{I}}_{2}}}{0.100\\text{L}}=1.1\\times {10}^{-3}M[\/latex]<\/li>\n<\/ol>\n

                                                                              14.\u00a0[latex]M=\\frac{\\text{mol}}{V}=\\frac{\\frac{1.0\\cancel{\\text{g}}}{40.08\\cancel{\\text{g}}{\\text{mol}}^{-1}}}{1.0\\text{L}}=0.025M[\/latex]<\/p>\n

                                                                              16.\u00a0[latex]\\begin{array}{c}\\frac{{C}_{1}{V}_{1}}{{C}_{2}}={V}_{2}\\\\ \\frac{\\frac{0.3556\\text{mol}}{\\text{L}}\\times 0.1718\\text{L}}{\\frac{0.1222\\text{mol}}{\\text{L}}}={V}_{2}\\\\ 0.5000\\text{L}={V}_{2}\\end{array}[\/latex]<\/p>\n

                                                                              18.\u00a0[latex]{V}_{1}=\\frac{{V}_{2}\\times {M}_{2}}{{M}_{2}}=25\\text{mL}\\times \\frac{0.025M}{0.33M}=1.9\\text{mL}[\/latex]<\/p>\n

                                                                              22. Determine the number of moles in 434.4 g of HCl: 1.00794 + 35.4527 = 36.4606 g\/mol<\/p>\n

                                                                              [latex]\\text{mol HCl}=\\frac{434.4\\cancel{g}}{36.4606\\cancel{g}{\\text{mol}}^{-1}}=11.91\\text{mol}[\/latex]<\/p>\n

                                                                              This HCl is present in 1.00 L, so the molarity is 11.9 M<\/em>.<\/p>\n

                                                                              24.\u00a0[latex]57\\text{g}{\\text{K}}_{2}{\\text{SO}}_{4}\\times \\frac{1\\text{mol}}{174.26\\text{g}}\\times \\frac{1\\text{L}}{0.20\\text{mol}}=1.6\\text{L}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                                                                              Part 2<\/p>\n

                                                                                \n
                                                                              1. Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO3<\/sub> by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3<\/sub> by mass?\n
                                                                                  \n
                                                                                1. Outline the steps necessary to answer the question.<\/li>\n
                                                                                2. Answer the question.<\/li>\n<\/ol>\n<\/li>\n
                                                                                3. What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?<\/li>\n
                                                                                4. What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g\/mL.<\/li>\n
                                                                                5. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm\u20133<\/sup> and contains 37.21% HCl by mass?<\/li>\n
                                                                                6. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of CaCO3<\/sub>, which is equivalent to milligrams of CaCO3<\/sub> per liter of water. What is the molar concentration of Ca2+ <\/sup>ions in a water sample with a hardness count of 175 mg CaCO3<\/sub>\/L?<\/li>\n
                                                                                7. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g\/mL and calculate the molarity of mercury in the stream.<\/li>\n
                                                                                8. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM<\/em> is observed, what is the concentration of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) in mg\/dL?<\/li>\n
                                                                                9. A cough syrup contains 5.0% ethyl alcohol, C2<\/sub>H5<\/sub>OH, by mass. If the density of the solution is 0.9928 g\/mL, determine the molarity of the alcohol in the cough syrup.<\/li>\n
                                                                                10. D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose (C6<\/sub>H12<\/sub>O6<\/sub>) in water. If the density of D5W is 1.029 g\/mL, calculate the molarity of dextrose in the solution.<\/li>\n
                                                                                11. Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H2<\/sub>SO4<\/sub>, for which the density is 1.3057 g\/mL.<\/li>\n<\/ol>\n
                                                                                  Show Selected Answers<\/span><\/p>\n
                                                                                  \n

                                                                                  1.\u00a0The answers are as follows:<\/p>\n

                                                                                    \n
                                                                                  1. The dilution equation can be used, appropriately modified to accommodate mass-based concentration units:
                                                                                    \n[latex]\\%{\\text{mass}}_{1}\\times {\\text{mass}}_{1}=\\%{\\text{mass}}_{2}\\times {\\text{mass}}_{2}[\/latex]
                                                                                    \nThis equation can be rearranged to isolate mass1<\/sub> and the given quantities substituted into this equation.<\/li>\n
                                                                                  2. [latex]{\\text{mass}}_{1}=\\frac{\\%{\\text{mass}}_{2}\\times {\\text{mass}}_{2}}{\\%{\\text{mass}}_{1}}=\\frac{10.0\\%\\times 400.0\\text{g}}{68.0\\%}=58.8\\text{g}[\/latex]<\/li>\n<\/ol>\n

                                                                                    3. The mass of the 10% solution is\u00a0[latex]1000{\\cancel{\\text{cm}}}^{3}\\times \\frac{1.109\\text{g}}{{\\cancel{\\text{cm}}}^{3}}=1.11\\times {10}^{3}\\text{g.}[\/latex]<\/p>\n

                                                                                    The mass of pure NaOH required is\u00a0[latex]\\text{mass}\\left(\\text{NaOH}\\right)=\\frac{10.0\\%}{100.0\\%}\\times 1.11\\times {10}^{3}\\text{g}=1.11\\times {10}^{2}\\text{g.}[\/latex]<\/p>\n

                                                                                    This mass of NaOH must come from the 97.0% solution:\u00a0[latex]\\begin{array}{l}\\\\ \\text{mass}\\left(\\text{NaOH solution}\\right)=\\frac{97.0\\%}{100.0\\%}=1.11\\times {10}^{2}\\text{g}\\\\ \\text{mass}\\left(\\text{NaOH solution}\\right)=\\frac{1.11\\times {10}^{2}\\text{g}}{0.970}=114\\text{g}\\end{array}[\/latex]<\/p>\n

                                                                                    5. \u00a0Since CaCO3<\/sub> contains 1 mol Ca2+<\/sup> per mol of CaCO3<\/sub>, the molar concentration of Ca2+<\/sup> equals the molarity of CaCO3<\/sub><\/p>\n

                                                                                    [latex]M{\\text{Ca}}^{2+}=\\frac{\\text{mol}{\\text{CaCO}}_{3}}{\\text{L}}=\\frac{175\\cancel{\\text{mg}}\\times \\left(\\frac{1\\text{mol}}{100.0792\\cancel{\\text{g}}}\\right)\\times \\left(\\frac{1\\cancel{\\text{g}}}{1000\\text{mg}}\\right)}{1\\text{L}}=1.75\\times {10}^{-3}M[\/latex]<\/p>\n

                                                                                    7. 1 mg\/dL = 0.01 g\/L and 1 L = 10 dL<\/p>\n

                                                                                    [latex]5.3\\cancel{\\text{mmol}}\\text{\/L}\\times 180.158\\text{mg\/}\\cancel{\\text{mmol}}=9.5\\times {10}^{2}\\text{mg\/L}[\/latex]
                                                                                    \n[latex]9.5\\times {10}^{2}\\text{mg\/L}\\times \\frac{1\\text{L}}{10\\text{dL}}=95\\text{mg\/dL}[\/latex]<\/p>\n

                                                                                    11. The molar mass of C6<\/sub>H12<\/sub>O6<\/sub> is [latex]6\\times 12.011+12\\times 1.00794+6\\times 15.9994=180.2\\text{g\/mol.}[\/latex] In 1.000 L, there are:<\/p>\n

                                                                                    [latex]\\begin{array}{l}\\left(1000\\cancel{\\text{mL}}\\times 1.029\\text{g}{\\cancel{\\text{mL}}}^{-1}\\right)=1029\\text{g}\\\\ \\text{mol dextrose}=1029\\cancel{\\text{g}}\\times 0.050\\times \\frac{1\\text{mol}}{180.2\\cancel{\\text{g}}}=0.29\\text{mol}\\end{array}[\/latex] C6<\/sub>H12<\/sub>O6<\/sub>.<\/p>\n

                                                                                    Since we selected the volume to be 1.00 L, the molarity of dextrose is\u00a0[latex]\\text{molarity}=\\frac{\\text{mol}}{\\text{L}}=\\frac{0.29\\text{mol}}{1.00\\text{L}}=0.29\\text{mol}[\/latex].<\/p>\n<\/div>\n<\/div>\n

                                                                                    Part 3<\/h3>\n
                                                                                      \n
                                                                                    1. Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.<\/li>\n
                                                                                    2. Consider the solutions presented:
                                                                                      \n\"In<\/p>\n
                                                                                        \n
                                                                                      1. Which of the following sketches best represents the ions in a solution of Fe(NO3<\/sub>)3<\/sub>(aq<\/em>)?<\/li>\n
                                                                                      2. Write a balanced chemical equation showing the products of the dissolution of Fe(NO3<\/sub>)3<\/sub>.<\/li>\n<\/ol>\n<\/li>\n
                                                                                      3. Compare the processes that occur when methanol (CH3<\/sub>OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.<\/li>\n
                                                                                      4. What is the expected electrical conductivity of the following solutions?\n
                                                                                          \n
                                                                                        1. NaOH(aq<\/em>)<\/li>\n
                                                                                        2. HCl(aq<\/em>)<\/li>\n
                                                                                        3. C6<\/sub>H12<\/sub>O6<\/sub>(aq<\/em>) (glucose)<\/li>\n
                                                                                        4. NH3<\/sub>(l<\/em>)<\/li>\n<\/ol>\n<\/li>\n
                                                                                        5. Why are most solid<\/em> ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid<\/em> (molten) ionic compound to be electrically conductive or nonconductive? Explain.<\/li>\n
                                                                                        6. Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:\n
                                                                                            \n
                                                                                          1. the solutions in Figure\u00a02<\/li>\n
                                                                                          2. methanol, CH3<\/sub>OH, dissolved in ethanol, C2<\/sub>H5<\/sub>OH<\/li>\n
                                                                                          3. methane, CH4<\/sub>, dissolved in benzene, C6<\/sub>H6\u00a0<\/sub><\/li>\n
                                                                                          4. the polar halocarbon CF2<\/sub>Cl2<\/sub> dissolved in the polar halocarbon CF2<\/sub>ClCFCl2\u00a0<\/sub><\/li>\n
                                                                                          5. O2<\/sub>(l<\/em>) in N2<\/sub>(l<\/em>)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                            Show Selected Answers<\/span><\/p>\n
                                                                                            \n

                                                                                            3. The answers are as follows:<\/p>\n

                                                                                              \n
                                                                                            1. Fe(NO3<\/sub>)3<\/sub> is a strong electrolyte, thus it should completely dissociate into Fe3+<\/sup> and (NO3<\/sub>\u2212<\/sup>) ions. Therefore, (z<\/em>) best represents the solution.<\/li>\n
                                                                                            2. Fe(NO3<\/sub>)3<\/sub>(s<\/em>) \u27f6 Fe3+<\/sup>(aq<\/em>) + 3NO3<\/sub>\u2212<\/sup>(aq<\/em>)<\/li>\n<\/ol>\n

                                                                                              5. The expected electrical conductivity is as follows:<\/p>\n

                                                                                                \n
                                                                                              1. high conductivity (solute is an ionic compound that will dissociate when dissolved);<\/li>\n
                                                                                              2. high conductivity (solute is a strong acid and will ionize completely when dissolved);<\/li>\n
                                                                                              3. nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water);<\/li>\n
                                                                                              4. low conductivity (solute is a weak base and will partially ionize when dissolved)<\/li>\n<\/ol>\n

                                                                                                7. The types of intermolecular attraction are as follows:<\/p>\n

                                                                                                  \n
                                                                                                1. ion-dipole;<\/li>\n
                                                                                                2. hydrogen bonds;<\/li>\n
                                                                                                3. dispersion forces;<\/li>\n
                                                                                                4. dipole-dipole attractions;<\/li>\n
                                                                                                5. dispersion forces<\/li>\n<\/ol>\n<\/div>\n<\/div>\n","protected":false},"author":57729,"menu_order":7,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1289","chapter","type-chapter","status-publish","hentry"],"part":877,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1289","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/users\/57729"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1289\/revisions"}],"predecessor-version":[{"id":1424,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1289\/revisions\/1424"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/parts\/877"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1289\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/media?parent=1289"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapter-type?post=1289"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/contributor?post=1289"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/license?post=1289"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}