Questions
Instructions to candidates
- 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 to read the question paper and make sure you have all the chemicals and apparatus that you may need.
- All working MUST be clearly shown.
- Mathematical tables and electronic calculators may be used.
- You are provided with:
- 0.1m sodium hydroxide solution F
- Solution G made by dissolving 9.0g of dibasic acid H2MO4 in 250cm3 of distilled water
You are required to:- Dilute solution G
- Standardize the diluted solution H using the sodium hydroxide solution F
- Determine the mass of M in the formula H2MO4
- Procedure I
- Using a measuring cylinder measure 20cm3 of solution G and transfer it into a beaker.
- Measure 80cm3 of distilled water and add it to the 20cm3 of solution G in the beaker. Label this as solution H. - Procedure II
- Place solution H in a burette. Pipette 25cm3 of solution F into 250cm3 conical flask. Add 2 – 3 drops of phenolphthalein indicator and Titrate with solution H. Record your results in table 1. Repeat the titration two more times and complete the table.- Table 1
I II III Final burette reading (cm3) Initial burette reading (cm3) Volume of solution H used (cm3) - Calculate the average volume of solution H used. ( 1 mark)
- Determine the number of moles of:-
- Sodium hydroxide in Solution F in 25cm3 ( 1 mark)
- Acid in solution H in the average volume used. ( 1 marks)
- acid in 100cm3 of solution H. ( 1 marks)
- acid in 20cm3 of solution G. ( 1 mark)
- acid in 250cm3 of solution G ( 2 marks)
- Calculate the:
- Molar mass of acid H2MO4 ( 2 marks)
- Mass of M in the formula H2MO4 given that H = 1, O=16. ( 1 marks)
- Table 1
- You are provided with:
- 0.15M ethan-1,2-dioc acid (oxalic), solution M
- 0.02M acidified potassium manganate (VII) solution N
You are required to determine the rate of reaction between acidified potassium manganate (VII) and ethan – 1,2 – dioc (oxalic) acid at different temperatures. - Procedure
- Place 5cm3 of solution N in a boiling tube
- Place another 5cm3 of solution M in another boiling tube.
- Heat solution N on a Bunsen burner flame to 800C. Allow it to cool to 700C.
- Add all solution M into solution N and at the same time start the stop watch
- Stir the mixture and record in table II the time taken for purple colour to disappear. At the same time record the temperature.
- Using clean boiling tubes repeat the procedure while allowing solution N to cool to 600, 500C and 450C in each case to complete table II below.
- Table II
Temperature before mixing °C 70 60 50 45 Temperature at which purple colour disappear °C Time taken for purple colour to disappear 1/time sec - On the grid of graph paper provided plot 1/time ( y-axis) against temperature at which the purple colour disappears. ( 3 marks)
- From your graph;
- determine the time taken for purple colour to disappear at 47.50C. ( 1 marks)
- State the relationship between rate of reaction and the temperature at which purple colour disappears. ( 1 mark)
- Table II
-
- You are provided with substance P for this question. Transfer the substance into a clean boiling tube. Add about 10cm3 of distilled water and stir. Pour the mixture into four clean test tubes of about 2cm3 each.
Observations Inferences
(1 mk)
(1 mk)- To the first portion of the solution, add sodium hydroxide solution dropwise until in excess.
Observations Inferences
(1 mk)
(½ mk) - Dip a clean stirring rod/glass rod/nichrome wire into the second portion and then place into the side of a blue bunsen flame.
Observations Inferences
(½ mk)(1 mk) - To the third portion, add 2-3 drops of barium nitrate solution followed by excess hydrochloric acid.
Observations Inferences
(1 mk)(½ mk) - To the fourth portion, add 2-3 drops of acidified potassium manganate (VII)
Observations Inferences
(½ mk)(1 mk)
- To the first portion of the solution, add sodium hydroxide solution dropwise until in excess.
- You are provided with solid K. Carry out the tests below. Write your observations and inferences in the spaces provided.
- Using a clean metallic spatula, heat about one third of solid K in a Bunsen burner flame.
Observations Inferences
(1 mk)(1 mk) - Dissolve the remaining portion of solid K into about 10cm3 of distilled water and divide the solution into 4 portions.
To the first portion, add two drops of acidified potassium permanganate solution.
Observations Inferences
(1 mk)
(1 mk) - To the second portion, add two drops of bromine water.
Observations Inferences
(1 mk)
(1 mk) - Determine the pH of the third portion using universal indicator paper.
Observations Inferences
(½ mk)(½ mk) - To the fourth portion, add a small amount of solid sodium hydrogen carbonate.
Observations Inferences
(1 mk)
(1 mk)
- Using a clean metallic spatula, heat about one third of solid K in a Bunsen burner flame.
- You are provided with substance P for this question. Transfer the substance into a clean boiling tube. Add about 10cm3 of distilled water and stir. Pour the mixture into four clean test tubes of about 2cm3 each.
Confidential
Requirements for candidates
In addition to the apparatus and fittings found in a Chemistry laboratory, each candidate will require the following.
- about 100cm3 of solution F
- about 50cm3 of solution G
- 30cm3 of solution M
- 30cm3 of solution N
- one burette 0 – 50ml
- one pipette 25ml
- two conical flasks
- 100ml measuring cylinder
- 200ml or 250ml beaker
- About 500ml distilled water
- Phenolphthalein indicator
- thermometer ( 0 – 1100C)
- Source of strong heat (preferably Bunsen burner)
- clock or stop watch
- 2 boiling tubes
- one CLEAN METALLIC spatula
- 6 clean dry test-tubes
- one test-tube holder
- at least 6cm length of universal indicator paper
- 0.5g of sodium hydrogen carbonate
- pH chart pH 1 – 14
- Bromine water supplied with a dropper
- 0.5g of solid K – oxalic acid.
- 0.5g of solid P – Sodium sulphite
The students should have access to the following
- 2.0M NaOH solution with a dropper
- 1.0M barium nitrate solution with a dropper
- Bromine water with a dropper
- Acidfied potassium manganate (vii) with a dropper
- 2.0M HCl with a dropper
- Bromine water is prepared by adding 1ml of liquid bromine to 100cm3 of distilled water and shaking thoroughly in a fume cupboard.
- Acidified potassium permanganate is prepared by adding 3.16g of solid potassium permanganate to 400cm3 of 2M sulphuric acid and diluting to one litre of solution using distilled water.
- Solution M is made by dissolving 12.6g of oxalic acid in 400cm3 distilled water and making it to 1 litre.
- Solution N is prepared by dissolving 3.16g of potassium manganate (VII) in 200cm3 of 2M sulphuric acid and adding more water to make 1 litre
- Solution F is prepared by dissolving 4g of sodium hydroxide pellets in about 800cm3 of distilled water and diluting it to one litre solution.
- Solution G is prepared by dissolving 9.0g of oxalic acid (ethan-1,2-dioic acid) in 200cm3 of distilled water and diluting it to 250cm3 solution.
Marking Scheme
-
- Complete table --------- 1mk
Must have 3 titrations done for 1mk
Penalise ½ mk once for any of the following
- Wrong arithmetic
- Inverted table
- Readings beyond 50cm3 unless explained
- Unrealistic titre value on the burette values below 1.0cm3 or above 100cm3
- Use of decimals – 1mk
Tied to 1st and 2nd rows only- Accept 1 or 2 dec. places used consistently
- If 2nd dec. place is used must be ‘O’ or ‘5’
(Penalise fully if any of the conditions is not met) Bdd
- Accuracy ---- 1mk
Compare the candidate reading to the school value
Conditions : (i) If any titre is within ± 0.1 of s.v 1mk- If none is within ± 0.1 of s.v but least within ± 0.2 s.v award 1mk
- If none is within ± 0.2 of s.v 0 mk
- Principle of Averaging 1mk
Conditions- If 3 consistent values are averaged 1mk
- If 3 titrations done and only 2 are possible and averaged 1mk
- If any 2 titrations are done inconsistent and averaged 0mk
- If 3 titrations are done, all are possible and only 2 averaged 0mk
- If 3 titrations are done are inconsistent and averaged 0mk
Penalties- Wrong Arithmetic i.e error outside ± 2 units in the 2nd dec. place penalise ½ mk
- If no work is shown but answer given is correct penalise ½ mk
- If the answer is rounded off to the 1st dec. place penalise ½ mk
- If no working is shown and answer given is wrong penalise fully - 0mk
- Final answer- 1mk
Compare to the s.v and tied to the correct average titre
Compare the candidates correct average titre with the s.v and- If within ± 0.1 of s.v ………………………….. 1mk
- If within ± 0.2 of s.v ………………………….. ½ mk
- If beyond ± 0.2 of s.v ………………………….. 0mk
Summary- CT - 1mk
- Dec –1mk
- AC- 1mk
- PA- 1mk
- FA- 1mk
05 - CALCULATIONS
-
- 25 x 0.1 = correct Ans
1000
Penalties- Penalise fully for strange figure
- Penalise ½ mk for wrong answer if error is outside ± 2 units in the 4th dec. place
- Accept answer given to at least 4 dec. places otherwise penalise ½ mk
- Units may not be shown, but if shown MUST be correct otherwise penalise ½ mk for wrong units
- mole ratio
NaOH : Acid (dibasic)
2 : 1 ½ ✓
∴ Answer I = corr. Ans
2 ✓ 1mk ✓ ½ mk
Penalties
Treat as in (i) - (iv) in CI above - 100 x Answer C(II) = correct answer
Titre volume ✓ ½ mk ✓½ mk
Penalties- Penalise ½ mk for WT (wrong transfer) of titre, otherwise penalise fully for strange figure
- Penalise ½ mk for wrong answer if the error is outside ± 2 units in the 4th dec place
- Treat as in (iii) – (iv) in C(i) above
- 20cm3 diluted to 100cm3 therefore number of moles in 20cm3 is equal to moles in 100cm3 = correct answer ✓ ½
Answer III same as IV
Penalties- penalise ½ mk for wrong Transfer (WT) otherwise fully for strange value
- Penalise ½ mk for rounding off answer to atleast 3 dec places
- Answer IV x 250 = correct answer
20 ✓½ mk 1mk
Penalties
Treat as in (i) – (iv) in C I above
- 25 x 0.1 = correct Ans
- Molar mass = 9.0 ✓ 1mk
Answer V
= correct Answer ✓1mk- Penalties
- penalise ½ mk for WT of answers in V, otherwise penalise fully for any strange figure used in the calculation
- Same conditions for units
- penalise ½ mk for not rounding off answer to a whole number
- H2MO4 = (2x1) + M+ ( 4 x 16) = Answer dI ✓ ½
= M + 66 = Answer d I 2 ✓ ½
M = Answer d I – 66
M = Correct answer✓ 1
Penalties- Penalise ½ mk for WT of answer in d II, otherwise penalise fully for any strange figure used in the calculation.
- Penalise ½ mk for no units given
- Penalise fully for answer if value of M is given as 38 or less
- Penalise fully for answer if value of M is as 60 or more.
Total 16 mks
Table marks 3mks distributed as
- Penalties
- Complete table --------- 1mk
-
-
- complete table with 12 correct readings 3mks
Incomplete table with 10 correct readings 2mks
Incomplete table with 8 correct readings 1 mk
Incomplete table with 6-7 correct readings ½ mk
Incomplete table with Less than 6 0mk
Conditions and penalties- Accept 1/t values to at least 3 d.p otherwise penalise ½ mk each to maximum of 1mk unless they work out exactly.
- Treat temp. reading ∠ 50° C in column II expt1 as unrealistic and penalise ½ mk once
- Penalise ½ mk for wrong units attached otherwise ignore if not stated.
- Use of decimals 1 mk
(Tied to temp. at which purple colour disappear and time taken only)- All readings of temp column II should either be whole nos or to 1 d.p. consistently for ½ mk otherwise penalise fully.
- All readings in column III for time should be either whole nos or to 2 d.p used consistently for ½ mk otherwise penalize fully.
- Accuracy 1mk
(Tied to 1st readings in column II and III only)- Temp. reading within ± 2°C of S.V should be credited ½ mk otherwise penalize fully
- Time reading within ± 5 seconds of school value should be credited ½ mk otherwise penalize fully.
- Trend 1mk
- Temp reading in column II should decrease across ½ mk
- Time reading in column III should increase across column ½ mk
- Penalise fully for any discrepancies in trends
- Use of decimals 1 mk
- complete table with 12 correct readings 3mks
- Graph 3mk distributed as
- Scale 1mk
- graph should cover atleast ½ of graph paper otherwise penalise fully
- Intervals should be uniform otherwise penalise fully
- Labelling – ½ mk
- Both axes should be labelled correctly
- Penalise fully for wrong units attached to axes otherwise ignore
- Plotting (1mk)
- Accept atleast 3-4 correct readings – 1mk
- 2 correct readings – ½ mk
- Less than 2 - 0mk
- Shape of curve ½ mk
- Accept shade if it is a line otherwise penalise fully
- Scale 1mk
- 1/t = correct reading at 47.5° C – ½ mk
- Time = 1 ✓½=✓ ans✓ 1mk
√ 1/t
- Time = 1 ✓½=✓ ans✓ 1mk
- 1/t = correct reading at 47.5° C – ½ mk
- Rate of reaction is directly proportion to temp. of reactants ✓1mk
½ mk ½ mk
-
-
Observations Inferences - Dissolve to form colourless solution - Soluble salt
- Absence of Fe2+, Fe3+, Cu2+a i Observations Inferences No white precipitate Na+, K+, NH4+ present
Pb2+, Al3+, Zn2+, Mg2+ absentii Observations Inferences Yellow flame Na+ present iii Observations Inferences White ppt that dissolve on adding HCl - CO2-3, SO2-3 present iv Observations Inferences Decolourises KMnO4 / turns acidified purple KMnO4 colourless SO2-3 present b i Observations Inferences K melts into a colourless liquid and Burns ½ mk with Smoky yellow flame ½ mk Accepts for 1mk
A long chain hydrocarbon
High carbon – hydrogen ration
- C ≡ C -Organic cpd tied to melting and burning unsaturated organic cpd ii Observations Inferences Acidified KMnO4 is decolourised✓ R – OH and or – C ≡ C-
OR it is a reducing agent ½ mk
Rej. Unsaturated hydro carboniii Observations Inferences Bromine water is decolourised
or – C ≡ C-
(1) penalise fully for any contradictory e.g. R-OH or RCOOH
(2.) Accept unsaturated cpd for ½ mkiv Observations Inferences pH 4 - 6 ✓ 1mk
Rej pH 76Weak acid ✓ 1mk
Accept for ½ mk, it is acidic / H+v Observations Inferences Effervescence / bubbles of gas✓1mk / hissing sound H+ or✓1mk - COOH
Acidic solution formed / carboxylic
acid / organic acid✓ ½ mk
Download Chemistry Paper 3 Questions and Answers with Confidentials - Moi Tea Mock Examinations 2022.
Tap Here to Download for 50/-
Get on WhatsApp for 50/-
Why download?
- ✔ To read offline at any time.
- ✔ To Print at your convenience
- ✔ Share Easily with Friends / Students