Page 180 - Theory and Problems of BEGINNING CHEMISTRY
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CHAP. 11]                               MOLARITY                                      169


               11.15. What is the concentration of a solution prepared by diluting 3.0 L of 2.5 M solution to 8.0 L with water?
                     Ans.  The number of moles of solute is not changed by addition of the water. The number of moles in the original
                           solution is
                                                             2.5 mol

                                                       3.0L         = 7.5 mol
                                                              1L
                           That 7.5 mol is now dissolved in 8.0 L, and its concentration is
                                                           7.5 mol
                                                                 = 0.94 M
                                                            8.0L
               11.16. What is the concentration of a solution prepared by diluting 0.500 L of 1.80 M solution to 1.50 L with
                     solvent?
                     Ans.  The number of moles of solute is not changed by addition of the solvent. The number of moles in the original
                           solution is
                                                            1.80 mol

                                                     0.500 L         = 0.900 mol
                                                              1L
                           That 0.900 mol is now dissolved in 1.50 L (compare Problem 11.17), and its concentration is
                                                          0.900 mol
                                                                  = 0.600 M
                                                           1.50 L
               11.17. What is the concentration of a solution prepared by diluting 0.500 L of 1.80 M solution with 1.50 L of
                     solvent?
                     Ans.  The number of moles of solute is not changed by addition of the solvent; it is 0.900 mol. What is the final
                           volume of the solution? If we add 1.50 L of solvent, it will be about 2.00 L. Compare this wording with that
                           of Problem 11.16. The 1.50 mol is now dissolved in 2.00 L, and its concentration is
                                                          0.900 mol
                                                                  = 0.450 M
                                                           2.00 L
               11.18. What is the concentration of a solution prepared by diluting 25.0 mL of 3.00 M solution to 60.0 mL?

                     Ans.  The number of moles of solute is not changed by addition of the solvent. The number of moles in the original
                           solution is
                                                    (0.0250 L)(3.00 mol/L) = 0.0750 mol
                           That 0.0750 mol is now dissolved in 0.0600 L, and its concentration is
                                                          0.0750 mol
                                                                  = 1.25 M
                                                          0.0600 L
                           Alternatively, the number of millimoles is given by
                                                   (25.0mL)(3.00 mmol/mL) = 75.0 mmol
                           The concentration is
                                                          75.0 mmol
                                                                  = 1.25 M
                                                           60.0mL
                           The use of millimoles and milliliters saves conversions of milliliters to liters.

               11.19. What concentration of salt is obtained by mixing 20.0 mL of 3.0 M salt solution with 30.0 mL of 2.0 M
                     salt solution?
                     Ans.  The final concentration is the total number of millimoles divided by the total number of milliliters. The
                           volume is 20.0 mL + 30.0 mL = 50.0 mL. The total number of millimoles is given by

                                         (20.0mL)(3.0 mmol/mL) + (30.0mL)(2.0 mmol/mL) = 120 mmol
                           The concentration is 120 mmol/50.0 mL = 2.4 M.
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