CHEMISTRY PAPER 3 Marking Scheme - 2017 MURANG'A MOCK EXAMINATION

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  1. Procedure Table 1 ( 5mks)
    1. Complete table   -   1mk
      Complete table with 3 titrations done – 1mk
      Incomplete table with 2 titrations done    ½ mk
      Penalties
      wrong subtraction
      Inverted table
      Burette readings beyond  50cm3  unless explained
      Unrealistic titre values eg less 1cm3  or in hundrends
      Pennalise  ½mk to  Maximum  ½mk

      Use of decimals – ( Tied to first and second rons  only) 1mk
      Accept one or two decimals used consistency otherwise penalize  fully ( A ward zero marks)
      Accept inconsistency  of use of zero’s on the initial burette readings eg   0, 0.0, 0.00.
      If two decimals, use the second digit as 0. Or 5 otherwise penalize fully.

      Accuracy – 1mk
      Compare the candidates titre values with the school value and award as follows.
      1. If at least one is within  t  0.1 of S.V – 1mk
      2. If none but at least one within t 0.2 of S.V – ½ mk
      3. If none within  t  0.2 of S.V award  0 mk.

      Principle of averaging -  1mk
      1. 3 Titrations done, 2 are consistent – 1mk
      2. If only 2 titrations done, are consistent – 1mk

      Final Accuracy  - 1mk
      tied to correct average of titre  - 1mk
      Compare the candidate correct average titre with S.V . award as follows.
      1. If  within  t 0.10cm3 of S.V  - 1mk
      2. If not  t  0.10  but within  t 0.20 of S.V   ½ mk
      3. If not within  T 0.20 cm3 of S.V 0mk

    2. 2NaOH + H2A → Na2A + 2H2 O
      2 moles  1 mole  NaOH 23 + 16 + 1 = 40
      Moles of NaOH  Used = ans
      Moles of dibasic acid in  25cm3 of solution M = ½  x  moles of NaOH  Above

    3.  
      1. 250cm3 of M= ans ( b) mole ratio 1: 2
        250cm3 of M  ½  X 0.0025 = ½  mk
      2. Answer in C(i) X 1000
        Answer (a) (average titre),  ½
      3. 6.3/answer in C(ii) ½  ans   ½
      4. r.f.m = 2 + 88 + 18n = ans  C( iii)
        18n = ans 

        Sample result

           F.B.R

        18.9

        18.8

        18.7

           T.B.R

        0.0

        0.0

         0.0

         Vol. of soln B used

        18.9

        18.8

        18.7



        Procedure II
        1. Complete table with 10 readings ( 3mks)               
          Incomplete table with 8  or 9  readings ( 2mks)
          Incomplete table with 5 or 6 readings ( 1mk
          Less than 2 readings  ( 0 –mk)
          Penalise ½ mk for incorrect 1/t   or value rounded to less than 3 d.p. unless exact ( maximum   penaly 1mk)

          Use of decimal (½  mk)
          should be whole number or 2 d.p consistent otherwise penalize fully.

          Accuracy – 1mk
          Compare the first record at 40o C of S.V  to candidates readings.
          H + 2o C award 1mk otherwise penalize fully

          Trend  -  ½  mk
          Time progression should be consistent or continuous drop from 40o   to 80o

          Graph -  3mks as shown
          Labelling of axes ( ½ mk)
          Scale  ( ½  mk)
          Area covered by the graph ( plots)  should be at least half of the grid provided.
          scale interval must be consistent on each  axis.
          Plotting – ( 1mk)
          3 – 4 [points correctly plotted award   -  1mk
          2 points are correctly plotted ½  mk
          Mark all points plots with a tick   or cross
          Line/ shape of graph- ( 1mk)
          Accept a correct line passing through at least 2 correctly- plotted point and origin. 1mk
        2. Calculation for  time at  65o C from graph time = Reciprocal of  1/t .  1mk

        3. the rate of reaction increases as the temperature increases due to K..e increases and collisions. 







          QUESTION 2

                           Observations

             Inferences

          a)Solid melts  ½   Red litmus paper ½  turns blue. Blue litmus paper  turns red. Colourless liquid seen in cooler part of test tube.

          NH4 + ions  ½   present / NH3  gas evolved.
          Acidic  ½  gas evolved  hydrarted salt.

          b) Solid Q dissolves to form Colourless solution ½

          Ionic compound / soluble salt /polar ½

          ii) White ppt is formed

          SO32-    SO32-   CL-   SO32-   present, 

          iii)  White ppt formed insoluble in acid

          SO4 2-  Pb2+  ,Al 3+  ions

          iv) White ppt formed 1 dissolves in excess to form colourless soln

          Zn2+  Pb2+ ,AL3+ ions present

          v)White ppt formed  Insoluble in excess  ½ mk

           Pb2+ , Al 3+  ions present

          vi) white ppt formed ½  dissolved on warming

           Pb2+ ions present  ½

          3.a) Soofty  flame burns with smoky flame ½

           

             b) Dissolve to form a colourless   generous solution

          Polar organic compound ½

          c.i) No observable change  1mk)

           R- OH  absent

            ii) Effectivescene / bubbles seen

           R –COOH  or

           iii) Purple acidified KMnO4 Decolourised ( Turns colourless )

           

 

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