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Observation                               Inference
                       - Effervescence/ fizzling/bubbles of a    Presence of – COOH/ H /H 3O  ions.
                                                                                         +
                                                                                               +
                       colourless gas.                           Solution is acid .    (1 mk)
          (ii)
                       - No effect on litmus paper.
                       Observation                             Inference
                       - The solution remained orange.         - Absence of R-OH.     (1 mk)

          (iii)
                    Observation                                Inference

                     Solution turns from purple to colourless      - Presence of  of –C C-
                    solution is decolourised     (1 mk)


                                                  SOIK DISTRICT

          1.     TABLE I
                 a)Complete table penalize ½ for  inverted table  and  arithmetic  errors
                                                      nd
                                               st
                 b) Use of decimal  tied to the 1  and 2  rows
                                        ½
                 c) Accuracy  ±0.2 s.v √    ± 0.1 sv√1
                 d) Principles of averaging  as shown below
                 e) Final answer  ± 0.2s.v  ± 0.1 s.v√1

                 a)T 1+T 2+T 3√
                              ½
                        3
                                    ½
                  = correct answer√   (2d.place)  (transferred  to the  table)
                 b)i)   5   √
                             ½
                           40
                        =0.125 moles  per litre
                                                                                            ½
                   ii)COOHCOOH (aq) + 2 NaH (aq)                       COONaCOONa (aq) +2H 2O (l) √   balanced
                                                                                            ½
                                                                                                                                                √  s.symbols
                                              OR
                    C 2H 2O 4(aq) +2 Na 2O 4(aq) +C 2Na 2O 4(aq)                       2H 2O(l)

                 iii) Moles of NaOH = 25X0.125 √
                                                      1000
                                            = 0.003125
                 Moles of C 2H 2O 4  = 0.003125 X 1
                                                           2
                                             = 0.0015625
                 Ans in (a)                       0.00015625

                           3
                 1000cm                      1000x0.0015625 √
                                                          ½
                                                        Ans in (a)
                               = Correct answer √
                                                  ½
                 V)    C 2H 2O 4 X H 2O  = answer in (iv)     √
                                                          ½

                        18x = Ans (iv) - 90√
                                            ½
                            x = Ans (iv) – 90 √
                                        ½
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