Chemistry Paper 3 Questions and Answers with Confidential - Joint Pre-Mock Exams 2021/2022

Instructions to Candidates

• Answer ALL questions 
• You are advised to take 15minutes at the beginning to read-through the question-paper very carefully and make sure you have ALL the chemicals and apparatus that you may need.
• Mathematical tables and electronic calculators may be used.
• All working must be clearly shown where necessary.

Q1.You are provided with:

1. Magnesium ribbon, solid FA1.
2. 0.7 M hydrochloric acid, solution FA2.
3. 0.05 M sodium hydroxide solution, solution FA3.
4. Distilled water.
5. Phenolphthalein indicator solution.
   You are required to determine the:
   1. Number of moles of hydrochloric acid that remain unreacted.
   2. Number of moles of magnesium that reacted.

      PROCEDURE I

1. Using a burette, measure 50.0cm³ of solution FA2 and place it in a 100ml plastic beaker.
2. Put the magnesium ribbon, solid FA1 in the 50.0cm³ of solution FA2 in the 100ml beaker and allow the reaction to proceed until effervescence stops.
   
   PROCEDURE II
3. Transfer ALL the solution obtained in Procedure I into a 250ml volumetric flask.
4. Top up the solution in the volumetric flask to calibration mark using distilled water. label this as solution FA4.
5. Empty the burette and fill it with solution FA3.
6. Pipette 25.0cm³ of solution FA4 and place it into an empty 250ml conical flask. Add 3 drops of the phenolphthalein indicator solution and titrate against solution FA3 from the burette. Record the results in Table 1.
7. Repeat the titration of solution FA4 against FA3 two more times in order to complete Table 1.
   Table 1.
   

   Titration	1st	2nd	3rd
   Final burette reading, cm3
   Initial burette reading, cm3
   Volume of solution FA3 used, cm3

   (4mks)
   Average volume of solution FA3 used = _______________________________ cm³ (1mk)
   Calculate the number of moles of:
   1. Sodium hydroxide in the average titre volume used. (1mk)
   2. Hydrochloric acid in 25.0 cm³ of solution FA4. (2mks)
   3. Hydrochloric acid in 250.0 cm³ of solution FA4. (2mks)
   4. Hydrochloric acid in 50.0 cm³ of solution FA2. (1mk)
   5. Hydrochloric acid that reacted with the magnesium. (2mks)
   6. Magnesium that reacted. (2mks)

Q2. You are provided with:

1. 2.20g of solid BA11,
2. Thermometer.
3. Distilled water.
4. 100 ml Glass beaker
5. Boiling tube.
6. You are required to determine the solubility of compound BA11 at various temperatures.

Procedure I:

1. Place the whole amount of solid BA11 supplied to you into a clean, dry boiling tube.
2. Using a burette, add 2.00cm³ of distilled water into the boiling tube with solid BA11.
3. Insert a thermometer into the boiling tube and heat the mixture in the hot water-bath (use the 100ml beaker), while stirring continuously with the thermometer, until the temperature of the mixture is about 800C when ALL the crystals JUST dissolve
4. Remove the boiling tube from the hot water bath and allow the contents to cool slowly while stirring with the thermometer. Note the temperature at which the crystals FIRST form/reappear and record this crystallization temperature, Tc in Table 2
5. Using the same mixture from (d) above, add 1.00cm³ of distilled water from the burette into the boiling tube containing the mixture and repeat steps (c) and (d) above. Continue in this way until a total volume of water added to the boiling tube is 7.00cm³. Complete Table 2 by calculating the solubility of compound BA11 in water at the different temperatures.
   Table 2: (4mks)
   

   Total volume of water added (cm3)	Crystallization Temperature ,TC(°C)	Solubility of compound BA11 in water (g/100g of   water)
   2.00
   3.00
   4.00
   5.00
   6.00
   7.00

6. On the grid provided plot a graph of solubility of compound BA11 (vertical axis) against temperature. (3mks)
7. Using the graph determine:
   1. the temperature at which 100.0g compound BA11 would dissolve in 100g of water. (1mk)
   2. solubility of compound BA11 at 30.0°C. (1mk)
8. A solution containing 100g of BA11 per 100g of water was cooled to 30.0°C. Determine the mass of crystals formed. (2mks)

Q3. You are provided with solids FA5, FA6 and FA7. Carry out the following tests and write your observations and interferences in the spaces provided.

1. Place all of solid FA5 in the boiling tube. Add about 10 cm³ of distilled water and shake until all the solid dissolves. Label this as Solution FA5.
   
   1. To about 2 cm³ of Solution FA5 in a test tube, add 2M sodium hydroxide solution drop wise until in excess.
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

   2. To about 2 cm³ of Solution FA5 in a test tube, add 2M ammonium hydroxide solution drop wise until in excess.
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

   3. To about 2 cm³ of Solution FA5 in the test tube, add 4 drops of 2M sulphuric (VI) acid.
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

   4. To about 2 cm³ of solution FA5 in a test tube, add 2 drops of potassium iodide solution..
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

2. Place solid FA7 into boiling tube. Add about 10cm³ of distilled water and shake well. Label this as solution FA7 .Use this solution for the following tests.
   1. Place about 2cm³ of solution FA7 in a test tube and place the universal indicator paper provided into the solution hence determine its pH.
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

   2. To about 2cm³ of solution FA7 made in (ii) above, add 3 drops of acidified potassium manganate (vii) solution.
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

   3. To the remaining solution FA7 in the boiling tube, add the other half of solid FA6.
      

      OBSERVATION	INFERENCES
      (1mk)	(1mk)

CONFIDENTIAL

QUESTION 1

1. Solid FA1, 3cm length of magnesium ribbon.
2. 60 cm³ of solution FA2 - 0.7M HCl.
3. 80 cm³ of solution FA3 - 0.05M NaOH.
4. 1 Burette .
5. One 25ml pipette.
6. One 100ml plastic beaker.
7. One wash bottle with distilled water.

QUESTION 2

1. 2.20g of solid BA11, (Oxalic acid)
2. Thermometer.
3. Distilled water.
4. 100 ml Glass beaker
5. Boiling tube.

QUESTION 3

1. 0.5g of solid FA5 - Pb (NO₃)₂.
2. 0.1g of solid FA7 - maleic acid.
3. 0.5g of solid FA6- Na₂CO₃.
4. Test tube – rack + 6 test tubes +1 boiling tube.
5. One metallic spatula.
6. Universal indicator paper

Access to:

1. Phenolphthalein indicator.
2. 0.5 M KI.
3. Acidified KMnO₄.
4. Universal indicator solution +pH chart (full range).
5. 2M H₂SO₄.
6. 2M NaOH.
7. 2M NH4OH
8. Bunsen burner.

 MARKING SCHEME

1. Table 1  (5mks)
   Award a maximum total of 5 marks distributed as follows.
   
   1. Complete table………………………1mk
      Conditions: either of the following:
      
      • Complete Table with 3 Titrations done………………1mk
      • Incomplete Table with only 2 Titrations done…………………½mk
      • Incomplete Table with only 1 Titration done…………0mrk
      • For no Titration done…………………………………0mk
        NOTE: where NO TITRATION done, penalize FULLY for ALL the marking points for Table 1.
        Penalties:
        before awarding a mark for complete Table; THE EXAMINER MUST ensure that none of the following mistakes is there in the Table- otherwise penalize ½mk for EACH MISTAKE but to a maximum penalty of ½mk(i.e. penalize ½mk ONCE even if there two or more mistakes):
      • Wrong arithmetic/subtraction
      • Inverted Table
      • Burette reading(s) beyond 50.0cm³, except where explained.
   2. Use of Decimals (tied to the 1^st and 2^nd ROWS ONLY)……………1mk
      Conditions:
      Either 1 or 2 decimal places used consistently.
      If 2 decimal places are used then the 2^nd decimal place MUST be either a “O”
   3. Accuracy (tied to CORRECT titre values ONLY)…………………1mk
      Compare the candidate’s CORRECT Titre values with the school values, (i.e. Teacher’s Average Titre):
      Conditions:
      
      • If at least one titre value is within ±0.10 cm³ of SV, then awarded……………………1mk
      • If No titre value is within ±0.10 cm³ of the SV but at least one titre value is within 0.20 cm³ of the SV, then award ½mark
      • If NONE of the titre values is within ±0.20 cm³ of the SV, then award………………0mark
        NOTE: If there is wrong arithmetic/ subtraction in the Table, compare the SV with the CORRECT worked out titre and award accordingly.
   4. Principles of averaging…………………………………1 mk
      Values averaged MUST be shown and MUST be consistent within0.20 cm³ of each other.
      Conditions.
      
      • If 3 consistent values are averaged………………………………….1mk
      • If 3 titrations are done and only 2 possible averaged………………..………1mk
      • If only 2 titrations are done, and are inconsistent and averaged………………1mk
      • If only 2Titrations are done, and are inconsistent and yet averaged……………………0mk
      • If 3 Titrations are done, ALL are possible and yet only 2 are averaged………… 0mk
      • If 3 inconsistent values are averaged…………………………0mk
      • If only 1 Titration is done………………0mk
        Penalties:
      • Wrong arithmetic, i.e. arithmetic error outside ±2 units in the 2^nd decimal place (e.g. 24.67 given as 24.64), penalize ½mk
      • If no WORKING is shown but “answer” given is correct, penalize ½mk.
      • If value is rounded off to the1st decimal place (e.g. 24.66 ≈ 24.7) or to a whole – number (e.g.24.33 ≈ 24), penalize ½mk.
      • If no working is shown but answer given is wrong, penalize FULLY, i.e. award 0mk
   5. Final accuracy (tied to the CORRECT average titre)…………………………….1mk
      
      1. Compare the candidate’s CORRECT average titre with the SV and award accordingly:
      2. If within ± 0.10 cm³ of the SV, award……………………………1mk
      3. If Not within ±0.10 cm³ of the SV, but it is within 0.20 cm3 of the SV, award ………..½mk
      4. If BEYOND ±0.20 cm³ of the SV, award……………………………0mk
         Table 1
         Complete table…………………….1mk
         Use of decimals……………………1mk
         Accuracy …………………………..1mk
         Principles of averaging……………1mk
         Final accuracy…………………….1mk
         Sub-total        5mks
         Calculations:
         1. Moles of NaOH in average titre = 0.05 ×30.0     √½mk
                                                                     1000
                                                                = 0.0015 mol √½mk
            Note:
            • Units may not be given but if given Must be correct otherwise penalize½mark foe wrong units attached to correct answer.
         2. Moles of HCl in 25 cm³ of solution FA4
            NaOH    :      HCl
                     1    :      1 √1mk
            0.0015 mol     0.0015 mol √1mk
         3. = 0.0015 × 250 √1mk =0.015 mol √1mk
                         25
            Note:
            • 0.0015 MUST be transferred and used INTACT otherwise penalize FULLY for any other figure used and award 0mk
            • Accept answer given to at least 4 decimals places, otherwise penalize ½mk for rounding off to 3 decimal places or less.
         4. Moles of HCl in 50 cm³ of FA2 = 0.7 × 50   √½mk
                                                                   1000
                                                             = 0.035 √½mk
         5. moles of HCl that reacted with magnesium = Ans in (v) – ans in (iv)
                                                                                    0.035 − 0.015√½mk = 0.02√½mk
         6. Mg : HCl
                1 : 2       √½mk
            = 0.02 ×1   √½mk = 0.01 mol √1mk
                     2
2. Table 2……………………4mks
   Award a MAXIMUM TOTAL of 4mks distributed as follows:
   1. Complete table………………………………………………….2mks
      Conditions/ penalties
      • Award ½mk for EACH experiment completely done.
      • Penalize ½mk for EACH solubility value either wrongly worked out or not work to a MAXIMUM penalty of 1mark.
      • Penalize 1mark ONCE for unrealistic Temperature readings, for any Temperature reading T≤25.0℃ and T>80.0℃.
      • Penalize½mk if ALL Temperature readings given in the Table are CONSTANT.
   2. Use of decimals(Tied to Temperature reading)………………………………… ½mk
      Accept ONLY if all readings recorded consistently either as whole numbers or to one decimal place of .0 or .5, otherwise penalize FULLY.
   3. Accuracy ………………………………………………………………………………………½mk
      Compare the candidate’s first Temperature reading (i.e. Temperature reading when the volume of water added is 5.00cm³) with the school value, SV (i.e. the Teacher’s temperature reading when the volume of water added is5.00cm³). If within ±2.0℃ of the SV, awarded 1mark, otherwise awarded 0mark.
   4. Trend ………………………………………………………………………………..1mk
      Award mark for Temperature readings showing a CONTINUOUS DROP, otherwise penalize FULLY.
      
      Hence the distribution of marks for table 2 is as follows.
      Complete table………………….2mks
      Use of decimals…………………½mk
      Accuracy ………………………….½mk
      Trend …………………………….…1mk
      Sub-total    4mks
      Answers  f,g and h.
   5. GRAPH………………………………………3mks
      Award MAXIMUM total of 3marks distributed as follows
      1. Labeling of the axes………………………………………½mk.
         Award ½mk ONLY if BOTH axes are CURRECTLY labelled.
         Penalties:
         • Penalize FULLY for inversion of axes
         • Penalize FULLY for wrong unit given, otherwise ignore if units are omitted.
         • Penalize FULLY if only ONE axis is correctly labelled.
      2. Scale ……………………………………………½mk
         Award ½mk for scale subject to the following conditions.
         Conditions:
         • Area covered by the PLOTS must be at least half the grid on EACH of the axis
         • Scale intervals MUST be consistent on EACH axis.
         • Scale chosen must be able to accommodate ALL plots/ points-Examiner MUST check the range of the readings on EACH axis.
           Note: panelize FULLY if any of the above three conclusions is NOT met.
      3. Plotting ………………………………………………1mk
         Award maximum of 1 mark for plotting.
         Conditions:
         • if 6 or 5 points are correctly plotted award ………….1mk
         • if only 4 or 3 points are correctly plotted award ½mk
         • If less than 3 points are correctly plotted award……….0 mark.
      4. Curve …………………………………………………………………………………1mk
         Award 1mark for a smooth rising curve joining at least three correctly plotted points.
         Note: Reject curve obtained by plotting 2 or more wrongly calculated solubility values from table 2.
         
         Hence the distribution of marks for the graph is as follows.
         Label of axes..……………………. ½mk
         Scale………………………………...…½mk
         Plotting………………………………..1mk
         Curve …………..………………………1mk
         Sub-total       3mks
   6.  
      1. Correct reading from the graph. √1mk
      2. Correct reading from the graph. √1mk
   7. 100g – ans in (g)(ii) √1mk
      Final ans √1mk
3.  
   1.  
      • 
        

        OBSERVATION	INFERENCES
        White ppt √½mk formed which then dissolves √½mk in excess alkali forming a colourless solution.                  (1mk)	Al2+, Pb2+, Zn2+,present√1mk All 3 ions given Only two ions given.½mk Only one ion given. 0mk (1mk)

      •  
        

        OBSERVATION	INFERENCES
        White ppt√½mk formed which is insoluble √½mk in excess aqueous ammonia               (1mk)	Al2+√½mk , Pb2+,√½mk present Penalize ½ mark for each contradictory ion.                (1mk)

      •  
        

        OBSERVATION	INFERENCES
        White ppt formed √1mk  Type equation here.                   (1mk)	Pb2+ present.√1mk                (1mk)

      •  
        

        OBSERVATION	INFERENCES
        Yellow White precipitates formed √1mk	Pb2+ present.√1mk Penalize FULLY for any contradictory ion.                 (1mk)

   2.  
      1. 
         

         OBSERVATION	INFERENCES
         Solution has pH=2 √1mk . NOTE. Reject pH given as range.                  (1mk)	Solution is strongly acidic. √1mk NOTE. Reject the solution is “strong acid”. Correct inference tied to correct pH.   (1mk)

      2.  
         

         OBSERVATION	INFERENCES
         KMnO4 solution is decolourised√1mk OR KMnO4 solution changes from purple to colourless.√1mk NOTE. Reject “ solution becomes/ turns colourless” or “ it turns colourless”	R-OH and >C=C</-C≡C- present √1mk

      3.  
         

         OBSERVATION	INFERENCES
         Effervescence producing a clourless gas √1mk	R-COOH present √1mk