Page 237 - Theory and Problems of BEGINNING CHEMISTRY
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226                                    SOLUTIONS                                 [CHAP. 15



               15.12. Calculate the mole fraction of the first component in each of the following solutions: (a)15.0gC 6 H 12 O 6
                     in 55.0 g H 2 O and (b)0.200 m solution of C 6 H 12 O 6 in water.

                     Ans.  (a)                             1 mol C 6 H 12 O 6
                                              15.0gC H 12 O 6           = 0.0833 mol C 6 H 12 O 6
                                                    6
                                                           180gC H 12 O 6
                                                                6

                                                             1 mol H 2 O
                                                   55.0gH O            = 3.06 mol H 2 O
                                                         2
                                                             18.0gH O
                                                                   2
                                                                0.0833 mol
                                                         =                   = 0.0265
                                                   X C 6 H 12 O 6
                                                           3.06 mol + 0.0833 mol
                           (b) In 1.00 kg H 2 O there are 0.200 mol C 6 H 12 O 6 and

                                                             1 mol H 2 O
                                                   1000 g H O          = 55.6 mol H 2 O
                                                          2
                                                             18.0gH O
                                                                   2
                               The mole fraction is
                                                                0.200 mol
                                                            =            = 0.00358
                                                      X C 6 H 12 O 6
                                                              55.8 mol total
               EQUIVALENTS
               15.13. An equivalent is defined as the amount of a substance that reacts with or produces 1 mol of what?
                     Ans.  In a redox reaction, electrons. In an acid-base reaction, hydrogen ions or hydroxide ions.
               15.14. What is the number of equivalents per mole of HCl?
                     Ans.  Assuming complete neutralization (or even oxidation of the Cl to Cl 2 ), we can see that 1 mol HCl will react
                                                                       −
                           with 1 mol NaOH (or 1 mol e ); thus, 1 equiv = 1 mol. (In its reactions to produce oxyanions of chlorine
                                                 −
                                    −
                           (ClO , ClO 2 , etc.), there is more than 1 equivalent per mole.)
                               −
               15.15. What numbers of equivalents per mole are possible for H 3 PO 4 in its acid-base reactions?
                     Ans.  1, 2, or 3, depending on how complete its reaction with base is.
               15.16. How many equivalents of NaOH react with the 82.0 g of H 2 SO 3 in part a of Example 15.9?
                                        −
                     Ans.  1 equiv. (One OH can react with 1 mol H ; 1 equiv of H 2 SO 3 is involved.)
                                                          +
               15.17. How many equivalents are there per mole of the first reactant in each of the following equations?

                     (a)Zn −→ Zn  2+  + 2e −                   (f ) NaHCO 3 + HCl −→ NaCl + CO 2 + H 2 O
                     (b)H 2 SO 4 + NaOH −→ NaHSO 4 + H 2 O     (g) NaH 2 PO 4 + 2 NaOH −→ Na 3 PO 4 + 2H 2 O
                     (c) 2 HCl + Ba(OH) 2 −→ BaCl 2 + 2H 2 O   (h)2 H 2 SO 4 + Cu −→ CuSO 4 + SO 2 + 2H 2 O
                              −      −     +        2+  + 4H 2 O (i)  NaHCO 3 + NaOH −→ Na 2 CO 3 + H 2 O
                     (d) MnO 4  + 5 e + 8H −→ Mn
                                     +
                     (e)H 2 SO 4 + 2H + Cu −→
                         Cu 2+  + SO 2 + 2H 2 O
                    Ans.  (a)2 (b)1 (c)1 (d)5 (e)2 (f)1 (g)2 (h) 2 [same as (e) for the one H 2 SO 4 molecule that is
                          reduced]  (i)1

               NORMALITY
               15.18. If a bottle is labeled 3.109 N H 3 PO 4 , which reaction should be assumed for the definition of its normality?
                     Ans.              H 3 PO 4 + 3 NaOH −→ Na 3 PO 4 + 3H 2 O  (complete neutralization)

               15.19. Explain why normality is the number of milliequivalents per milliliter.
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