INSTRUCTIONS TO CANDIDATES
- Answer ALL the questions
- You are not allowed to start working with the apparatus for the first 15 minutes of the 2 ¼ hours allowed for this paper. This time is to enable you read the question paper and make sure you have all the chemicals and apparatus that you may need.
- Mathematical tables and silent electronic calculators may be used
- ALL workings MUST be clearly shown where necessary.
FOR EXAMINER’SUSE ONLY
QUESTION |
MAXIMUM SCORE |
CANDIDATE’S SCORE |
1 |
23 |
|
2 |
10 |
|
3 |
7 |
|
TOTAL SCORE |
40 |
QUESTIONS
-
- You are provided with;
- 0.3 g of metal A
- 70cm3 of 1.0 M hydrochloric acid solution labelled as solution B.
- 100cm3 of 0.1 M sodium hydroxide labeled solution C.
- Phenolphthalein indicator.
- You are provided with;
You are required to determine the relative atomic mass of metal A
Procedure I
- Using a burette measure 50cm3 of solution B into a 250ml beaker.
- Add the whole amount of solid A provided into the beaker containing 50.0cm3 of solution B and swirl carefully until ALL the solid reacts completely.
- Transfer the mixture left in the beaker after the reaction into a 250ml volumetric flask. Rinse the beaker with distilled water and transfer all the rinsing’s into the volumetric flask. Make up the volume of the solution in the volumetric flask up to the mark with distilled water, shake well and label the solution D.
- Fill a clean burette with solution D.
- Pipette 25cm3 of solution C into a 250ml conical flask, add 3 drops of phenolphthalein indicator solution and titrate against solution D from the burette. Record your results in table I below.
- Repeat the titration TWO more times to complete table I.
Table I (4marks)
I |
II |
III |
|
Final burette reading, cm3 |
|||
Initial burette reading, cm3 |
|||
Volume of solution D used, cm3 |
- Calculate the average volume of solution D used, cm3 (1mark)
- Calculate
- Calculate the number of moles of HCI in 50.0 cm3 of solution B. (1mark)
- Determine the number of moles of NaOH in 25.0 cm3 of solution C. (1mark)
- Determine the number of moles of HCl in the average volume of solution D used in the titration. (1mark)
- Calculate;
- The number of moles of HCl in 250cm3 of HCI solution D. (1mark)
- The number of moles of HCl that reacted with metal A. (1 mark)
- Given that metal A forms a divalent cation,
- Determine the moles of metal A that reacted with hydrochloric acid. (1mark)
- Determine the Relative atomic mass of metal A. (1 mark)
1. (b) You are provided with
- 2.0g of solid E in a dry boiling tube.
- A thermometer
- Distilled water
- Hot water bath
You are required to determine the temperatures at which solutions of known concentrations of compound K becomes saturated and plot a solubility curve.
Procedure II
- Using a burette, add 5.0 cm3 of distilled water into the boiling tube with solid E.
- Place the boiling tube into a water bath and warm it while stirring with a thermometer until all the solid dissolves.
- Remove the boiling tube from the hot water and allow the content to cool slowly while stirring with the thermometer. NOTE the temperature at which crystals start to appear and record this temperature in table II.
- Add a further 2.0 cm3 of distilled water from the burette into the boiling tube containing the mixture and repeat steps (c) and (d) above.
- Repeat the procedure (e) above until the volume of water added is 15.0 cm3.
- Complete table II by calculating the solubility of compound E in water at different temperatures.
Table II (5 marks)
Total volume of water added (cm3) |
Temperature at which crystals first appear. (oC) |
Solubility of substance E in g/100g water. |
5.0 |
||
7.0 |
||
9.0 |
||
11.0 |
||
13.0 |
On the grid provided plot a graph of solubility of compound E (vertical axis) against temperature. (3 marks)
- From the graph determine the temperature at which the solubility of E in water at 35.0 ºC/100g of water (1mark)
- Determine the mass of crystals formed when a hot saturated solution of compound E is cooled from 70ºC to 45ºC (1 mark)
- What is the relationship between the solubility of compound E and change in temperature (1mk)
2. You are provided with solid F. Carry out the tests below and record your observations and inferences in the spaces provided.
Place all the solid F provided in a boiling tube. Add 10 cm3 of distilled water and shake well. Divide the resulting solution into five portions of 2 cm3 each.
Observations |
Inferences |
(1 mark) |
(1mark) |
To the first portion, add 2 drops of barium nitrate solution followed by 2 cm3 of dilute nitric (V) acid.
Observations |
Inferences |
(1mark) |
(½mark) |
To the second portion, add 2-3 drops of lead (II) nitrate solution followed by 2 cm3 of dilute nitric (V) acid.
Observations |
Inferences |
(1mark) |
(½mark) |
To the third portion add about 2-3 drops of sodium sulphate.
Observations |
Inferences |
(1mark) |
(1 mark) |
To the 4th portion add sodium hydroxide dropwise until in excess.
Observations |
Inferences |
(1mark) |
(½ mark) |
To the 5th portion add ammonia solution dropwise until in excess.
Observations |
Inferences |
(1mark) |
(½ mark) |
3. You are provided with solid G. Carry out the tests below and record your observations and inferences in the spaces provided.
Scoop a little of solid G using a metallic spatula and burn it in a Bunsen burner flame.
Observations |
Inferences |
(1mark) |
(1mark) |
To the remaining portion of solid G, add about 6cm3 of distilled water and shake. Divide the mixture into two portions.
Observations |
Inferences |
(1mark) |
(1mark) |
To the first portion add 2-3 drops of acidified potassium manganate (VII) solution.
Observations |
Inferences |
(1mark) |
(1mark) |
To the second portion add solid sodium carbonate provided.
Observations |
Inferences |
(½mark) |
(½mark) |
MARKING SCHEME
- TABLE I……………..Total 5 mks distributed as below
- Complete table ----------------------------1mk
Complete table with 3 titrations ------1mk
Penalties
Unrealistic titre values i.e. values below 1cm3 or hundreds
Burette readings beyond 50cm3 unless explained
Inverted table
Wrong arithmetic
N/B: Penalize ½ mk each for a maximum of ½ mk - Decimals ----------1mk(Tied to the 1st and 2nd rows only)
Should be 1 decimal place or 2 decimal places used consistently otherwise penalise fully.
The 2nd decimal place should either be a 0 or 5 - Accuracy ------------1mk
If any titre value within ± 0.1 of the school value ---------------------------------award 1mk
If any titre within ± 0.2 of the school value ½ mk
If none of the titre values is within ± 0.2 of school value (s.v)-----------award 0mk - Principles of Averaging -------------1mk
Values to be averaged MUST be shown and MUST be within ± 0.2 of each other
Conditions
If 3 consistent values are averaged -----------1mk
3 titrations done only 2 are possible and averaged ---1mk
3 consistent values but only 2 are averaged award 0mk
3 inconsistent values are average award 0mk
Penalties
Penalise ½ mk for arithmetic error in answer outside ± 2 units in the 2nddec. Place.
Penalise ½ mk for NO WORKING shown but the answer is correct.
Accept rounding off to the 2nddec.places otherwise penalise ½ mk if answer is rounded off to 1stdec. place unless values divide exactly to 1 dec. place - Final answer --------------------1mk
(Compared to school value (S.V) Tied to correct average titre)
If within ± 0.1 of S.V --------- award 1mk
If within ± 0.2 of S.V -------- award ½mk
If beyond ± 0.2 of S.V--------- award 0mk
- Complete table ----------------------------1mk
-
- Moles of HCI in 50cm3 of solution B
= (1.0 ×50 )/1000㇢½=0.05 moles㇢½ - Moles of NaOH in 25cm3 of solution C
= (0.1×25)/1000㇢½=0.0025 moles㇢½ - Moles of HCI in average volume of solution D used
Mole ratio of NaOH: HCI = 1: 1㇢½
Therefore moles of HCI = moles of NaOH = 0.0025 moles. ㇢½
- Moles of HCI in 50cm3 of solution B
-
- Moles of HCI in 250 cm3 of solution D.
(0.0025 ×250 )/(average volume )㇢½=correct answer.㇢½ - Moles of HCI that reacted with metal A.
= 0.05 - answer c (i) above ㇢½
= correct answer ㇢½
- Moles of HCI in 250 cm3 of solution D.
-
- Moles of A reacted:
A + 2HCI → ACI2 + H2
= answer c (ii) above ÷ 2 ㇢½
= correct answer ㇢½ - Relative atomic mass of metal A
= (0.3 )/(answer d(i) above )㇢½
=correct answer.㇢½ (Reject answer in decimal places)
- Moles of A reacted:
PROCEDURE II
TABLE II ……………..Total 5 mks distributed as below
- Temperature column------------------------------------------------------------------------2½ mks)
Completely filled table ……….. 1 mk
decimals ………………………..1 mk
(Accept whole numbers or 1 d.p (.5) or 2 d.p (.25, 50 or .75)
Trend -----------------------------------½ mk
(Should be a continuous decrease in temperature with increase in volume of water added) - Solubility values------------------------------------------------------------------------------- 2½ mks)
(Award ½ mk for each correctly worked out value of solubility)
NOTE:
Penalize ½ mk each for wrongly worked out value of solubility up to a maximum of 2½ mks - Graph.
Labeling of axes (both) ----------------------------------------------------------------------------- ½ mk)
Penalise ½ mk for wrong units used in any of the axis otherwise ignore if units not given.
Penalise ½ mk for inverted axis
Accept for ½ mk if no units shown on labeling.
Scale ----------------------------------------------------------------------------------------------------½mk)
Area covered by actual plots must be at least half of the big squares (y-axis) and half of the big squares ( x – axis)otherwise give zero.
Scale used must be consistent on both axes, otherwise penalise fully.
Plotting -------------------------------------------------------------------------------------------------1mk)
Accept 5 points correctly plotted for 1mk
If 3 or 4 points are correctly plotted award ½ mk
If less than 3 are correctly plotted award 0mk
Accept correct plots even if the axes are interchanged.
Curve -----------------------------------------------------------------------------------------------------1mk)- Award ½mk for extrapolation
Award another ½ mark for correct reading.
Penalize ½ mark for wrong units or no units given. - Correct solubility reading at 70ºC – correct solubility reading at 45ºC ㇢½
= correct answer ㇢½ (Penalize ½ mark for wrong units or no units given) - Solubility of compound E increases with increase in temperature.
- Award ½mk for extrapolation
QUESTION 2
Observations |
Inferences |
Solid dissolves ㇢½ to form a colourless solution㇢½ |
Salt is soluble ㇢1 Accept for ½ mk – Cu2+, Fe2+, Fe3+ absent |
Observations |
Inferences |
White precipitate ㇢½ that does not dissolve ㇢½ in nitric (v) acid. |
SO42- present ㇢½ (Penalize fully for any contradictory ion/s) |
Observations |
Inferences |
White precipitate ㇢½ that does not dissolve ㇢½ in nitric (v) acid. |
SO42- confirmed㇢½ (Penalize fully for any contradictory ion given) |
Observations |
Inferences |
No white precipitate㇢1 Accept for ½ mk - No observable change - Solution remains colourless Reject No change. No reaction. i.e award zero mk. |
K+, Na+ , Mg2+, Zn2+, Al3+ present All five given -award 1 mk. Any 3 correct – award ½ mk Less than 3 given -award 0 mk. Penalize ½ mk each for any contradictory ion up to a maximium of 1 mk. Accept for ½ mk Ca2+,Ba2+,Pb2+ absent s |
Observations |
Inferences |
White precipitate㇢½ formed insoluble㇢½ in excess |
Mg2+ present ㇢½ |
(vi)
Observations |
Inferences |
White precipitate㇢½ formed insoluble㇢½ in excess |
Mg2+ present ㇢½ |
QUESTION 3
Burning a little solid G using a Bunsen burner flame.
Observations |
Inferences |
Burns with a yellow㇢½ sooty ㇢½ flame – accept burns with a luminous flame for ㇢1 |
=C=C= / -C≡C- present㇢1 (for either) |
Add about 6cm3 of distilled water and shake.
Observations |
Inferences |
Solid dissolves to form a colourless solution. ㇢½ |
Solid is polar/ polar compound. ㇢½ |
To the 1st portion add acidified potassium manganate (VII) solution.
Observations |
Inferences |
Purple acidified potassium manganate (VII) is decolourised. ㇢1 / changes to colourless |
=C=C= / -C≡C- present ㇢½ (for either) R-OH ㇢½ |
(d) To the 2nd portion add sodium hydrogen carbonate
Observations |
Inferences |
Effervescence /bubbles /fizzing㇢½ Reject: fissing/fizzling/hissing. |
R-COOH㇢½ or / H+ /H3O+ - Reject fully any contradictory functional group. i.e. award 0 mk. |
CONFIDENTIAL
The information contained in this confidential is to enable the head of institution and the teacher in charge of chemistry to make adequate preparations for the practical examination.
Note: No one else should have access to this information either directly or indirectly
INSTRUCTIONS
In addition to the laboratory fittings each candidate requires the following:
- Solid E – potassium chlorate 2.0 g accurately weighed in a dry boiling tube.
- Solid A – 0.3 g magnesium powder
- 70 cm3 of Solution B- 1M HCI
- 120 cm3 of Solution C – 0.1M NaOH
- 2 conical flasks.
- Solid G – 0.2 g of maleic acid in a stoppered container
- Solid F- 0.2g of magnesium sulphate.
- 2 g of solid Na2CO3.
- A burette (50ml)
- A pipette (25ml) and pipette filler
- Thermometer (-10oC – 110oC)
- Volumetric flask(250ml)
- Complete retort stand
- A white tile
- Six test tubes in a rack
- One boiling tube
- Test tube holder
- 10ml measuring cylinder
- 500ml distilled water in a wash bottle
- One metallic spatula
- one label
Access to:
- Source of heat
- Water bath
- 2M NaOH solution
- 2M dilute nitric (V) acid
- 2M aqueous ammonia
- 1M barium nitrate
- 1M lead (II) nitrate
- 1M sodium sulphate supplied in a dropper bottle
- Acidified potassium manganate (VII) solution
- Phenolphthalein indicator.
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