Page 231 - Theory and Problems of BEGINNING CHEMISTRY
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220                                    SOLUTIONS                                 [CHAP. 15


               a solution. The symbol for molality is m. Its unit is molal. Carefully note the differences between molality and
               molarity:
               1. Molality is defined in terms of kilograms, not liters.
               2. Molality is defined in terms of solvent, not solution.
               3. The symbol for molality is a lowercase m, not capital M.
               Take great care not to confuse molality and molarity.

               EXAMPLE 15.3. Calculate the molality of a solution prepared by adding 0.300 mol of solute to 225 g of water.
                                                      0.300 mol
               Ans.                                           = 1.33 m
                                                       0.225 kg
               EXAMPLE 15.4. What mass of H 2 O is required to make a 2.50 m solution with 0.511 mol NaCl?
                                                         1.00 kg H O

               Ans.                        0.511 mol NaCl       2    = 0.204 kg H O
                                                        2.50 mol NaCl          2
               EXAMPLE 15.5. What mass of 1.25 m CH 3 OH solution in water can be prepared with 50.0 g of CH 3 OH?
                                                1 mol CH 3 OH  1.00 kg solvent

               Ans.                 50.0gCH OH                              = 1.25 kg solvent
                                           3
                                                32.0gCH OH   1.25 mol CH 3 OH
                                                       3
                                                      50.0gCH OH + 1250 g H O = 1300 g solution
                                                             3           2
                     Note the importance of keeping track of what materials you are considering, in addition to their units.

               EXAMPLE 15.6. Calculate the molarity of a 1.85 m solution of NaCl in water if the density of the solution is 1.04 g/mL.
               Ans.  Since neither concentration depends on the quantity of solution under consideration, let us work with a solution
                     containing 1.000 kg H 2 O. Then we have

                                                          58.5 g NaCl
                                              1.85 mol NaCl          = 108 g NaCl
                                                          1 mol NaCl
                     The total mass of the solution is therefore 1108 g, and its volume is

                                                       1mL
                                                1108 g       = 1070 mL = 1.07 L
                                                       1.04 g
                     The molarity is
                                                     1.85 mol NaCl
                                                                 = 1.73 M
                                                        1.07 L


               15.3. MOLE FRACTION
                   The other temperature-independent concentration unit introduced in this chapter is the mole fraction. The
               mole fraction of a substance in a solution is the ratio of the number of moles of that substance to the total number
               of moles in the solution. The symbol for mole fraction of A is usually X A , although some texts use the symbol
               N A . Thus, for a solution containing x mol of A, y mol of B, and z mol of C, the mole fraction of A is
                                                           x mol A
                                            X A =
                                                  x mol A + y mol B + z mol C
               EXAMPLE 15.7. What is the mole fraction of CH 3 OH in a solution of 12.0 g CH 3 OH and 25.0 g H 2 O?

                                                      1 mol CH 3 OH
               Ans.                       12.0gCH OH               = 0.375 mol CH 3 OH
                                                  3
                                                      32.0gCH OH
                                                              3
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