Notes to award marks on the table. TABLE 1
| Titre number | I | II | CT = 1 DP = 1 Acc = 1 Ap = 1 FA = 1 5 III |
| Final burette reading cm3 | 28.0 | 28.0 | 28.0 |
| Initial burette reading cm3 | 0.0 | 0.0 | 0.0 |
| Titre volume cm3 | 28.0 | 28.0 | 28.0 |
- CT – complete table 1mk
The table should be filled completely
Errors like
- Values recorded beyond 50.0cm3
- wrong arithmetic between final and initial
- inversion of the table
- titre values less than 1
penalize ½ mk only for any point of the errors mentioned above.
- D.p. = Decimal point 1mk
Either all values recorded to 1 d.p OR all recorded to 2 d.p, second figure of the d.p being 0 or 5 only e.g. 23.60 , 24.75 not 21.36 , 22.57
A student should not record same values to 1 d.p other to 2 d.p in the same table. If this happens award 0mk for d.p
- Acc. = Accuracy …………1mk
Consider any one of the candidates’ titre if within 0.0 to 0.1cm3 of school value (s.v) award 1 mk
±0.11 to ± 0.20cm3 of school value (s.v) award ½ mk
If beyond ± 0.20cm3 award 0mk
- PA = Principles of Averaging ……1mk
i) Candidates to average 3titres if they are within ± 0.1cm3 to one another i.e. maximum deviation between smallest and largest being 0.20cm3, if beyond this limit award 0mk
Or
- Candidates to average 2 titres if only 2 titres are possible
- If a candidate averages 2 titres when three are possible award 0mk for averaging.
- Averaged titre to be recorded to 2d.p if it is recurring e.g. 26.333 to 26.33 not 26.3 or 26.35
Round off or truncate allowed e.g 24.6666 to 24.67 or 24.66
- F.A – Final Answer ………………1mk
This is correctly averaged titre compared to school value.
Award marks as for accuracy.
0.0 to ± 0.10 ½ mk
0.11 to ± 0.20 ½ mk
Summary of award of marks to be written beside the table
N/B
School value is obtained by the teacher performing the experiment and calculating the average titre.
CALCULATIONS PROCEDURE 1
- Molarity of solution A = 6.95 x 1000
278 250
= 0.1 molar ½ mk
No. of moles = 0.1 x 25 = 0.0025moles ½ mk
1000
- Mole ration A:B is 5 : 1 ½ mk
So moles of B = 1/5moles of A
= 1/5 x 0.0025 = 0.0005moles ½ mk
- Av. Titre had 0.005mol.
Therefore 100cm3 of solution B has = 1000 x 0.005
Average titre ½ mk
e.g. 1000 x 0.0005 = 0.0179
28 NB: use candidate’s titre value in part (i) above
PROCEDURE II
TABLE II 5mks
Award marks as shown in table I
Calculations Procedure II
- 1000cm3 of solution B has
Therefore Average titre has
= Av, Titre table II x (ans. in c above) ½ mk
1000
= ans ½ mk
- Mole ration B: C is 2:5
No. of mole of C = 5/2 moles of B ……………… ½ mk
= 5/2 x ans. in (a) above.
e.g. 5/2 x 0.00429 = 0.001073 ½ mk
- 25cm3 of solution. C has (ans in (b) above)
1000cm3 of solution C. has 1000 x ans. (b) above ½ mk
25
= ans.
e.g. 24 x 0.0179 = 0.000429
1000 ½ mk
Grand total marks for Q1 = 17mks
Question 2
Table III
| Volume of water in boiling tube cm3 | Temp. at which crystals of F first appear 0C | CT2 D1 AC ½ Td ½ Sol 2 6 Solubility of solid F (g/100g of water) |
| 15 | 80 | 26.5 |
| 20 | 71 | 20 |
| 25 | 65 | 16 |
| 30 | 52 | 13.3 |
| 35 | 48 | 11.4 |
| 40 | 45 | 10 |
Notes on marking table
CT =Complete table …………….2mk (temp. column only)
2mks for all temp. values filled, 1 ½ mkfor only 4 or 5 entries made, 1mk for 3 entries and 0mk for only 2 or 1 values entered
Penalize ½ mk for all temp. readings above 84.50 and below 100C to a maximum of 1mk
D.P = Decimal point ………1mk
All temp. values to be recorded as whole numbers OR with a decimal as 0 or 0.5 only if any other figure is used award 0 mk for d.p
Accuracy ……… ½ mk
Compare the candidates temp. reading at volume 15cm3 with that of the teacher.
If ±20C from the school value award ½ mk
Trend ………………. ½ mk
Award ½ mk for continuous temp. drop, otherwise penalize fully.
Solubility column …………….. 2mks
For 6 correct entries ……………………2mks
For 4 or 5 correct entries …………….. 1 ½ mks
For 3 correct entries …………………. 1mk
Below 3 entries ………………………0mk
- Graph …………………3mks
Labelled axis ……………… ½ mk
Axis to be labeled with quantities and unit i.e. Temp in 0C and solubility ( g / 100g of water).
Scale ………………. ½ mk
Plot to cover 2/3 of the given grid.
Plots …………..1mk
To be plotted accurately
Curve ……………….1mk
Plots to be joined to give a smooth curve increasing with increase in temp. award 1mk smooth curve passing through any experimental values one MUST be through 26.5g/100g of water.
- Showing on graph at solubility 15g/100g of water (only from a correct curve) …….½ mk correct reading…….1/2mark
Total marks for Question 2 = 10mks
Question 3. (13mks)
| Observation | Inference |
| Colourless liquid forms on cooler sides of tests tubeColourless gas produced which relights a glowing splintBrown gas with a pungent smell produced Each ½ mk to a max. of 1mk | Solid D is hydrated ½ mkNO3– ions present ½ mk |
| Observation | Inference |
| White ppt ½ mk Soluble in excess ½ mk | AL3+, Zn2+, or Pb2+ present 1mk for 3 cations½ mk for 2 cations0mk for 1 cation |
| White ppt ½ mk Insoluble in excess ½ mk | Al3+ ½ mk or Pb2+ ½ mk present |
| No white ppt formed 1mk` | Al3+, present 1mk Or Pb2+ absent ½ mk |
| Burns with a blue smokeless flame ½ mk | Saturated organic compound Or Hydrocarbon with low C:H ratio {any point ½ mk} or C C |
| Purple KMnO4 turns colourless 1mk | |
Orange K2Cr2O7 turns green 1mk | R – OH present 1mk |
Yes