- Procedure Table 1 ( 5mks)
- 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.- If at least one is within t 0.1 of S.V – 1mk
- If none but at least one within t 0.2 of S.V – ½ mk
- If none within t 0.2 of S.V award 0 mk.
Principle of averaging - 1mk- 3 Titrations done, 2 are consistent – 1mk
- If only 2 titrations done, are consistent – 1mk
tied to correct average of titre - 1mk
Compare the candidate correct average titre with S.V . award as follows.- If within t 0.10cm3 of S.V - 1mk
- If not t 0.10 but within t 0.20 of S.V ½ mk
- If not within T 0.20 cm3 of S.V 0mk
- 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 -
- 250cm3 of M= ans ( b) mole ratio 1: 2
250cm3 of M ½ X 0.0025 = ½ mk - Answer in C(i) X 1000
Answer (a) (average titre), ½ - 6.3/answer in C(ii) ½ ans ½
- 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- 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 - Calculation for time at 65o C from graph time = Reciprocal of 1/t . 1mk
- 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 )
- Complete table with 10 readings ( 3mks)
- 250cm3 of M= ans ( b) mole ratio 1: 2
- Complete table - 1mk
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