Chemistry Paper 3 Questions and Answers with Confidential - Form 3 End Term 3 Exams 2021

1. You are provided with:
   • 0.1m sodium hydroxide solution F
   • Solution G made by dissolving 23.46g of dibasic acid H₂MO₆ in 250cm³ of distilled water
     You are required to:
     1. Dilute solution G
     2. Standardize the diluted solution H using the sodium hydroxide solution F
     3. Determine the mass of M in the formula H₂MO₆
        Procedure 1
        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.
        1. Table 1
           

           	I	II	III
           Final burette reading (cm3)
           Initial burette reading (cm3)
           Volume of solution H used (cm3)

           (4 marks)
        2. Calculate the average volume of solution H used. (2 mark)
        3. Determine the number of moles of:-
           1. Solution F in 25cm³ (3 mark)
           2. Acid in solution H in the average volume used. ( 2 marks)
           3. Acid in 100cm³ of solution H. ( 2 marks)
           4. Acid in 20cm³ of solution G. ( 2 mark)
           5. Acid in 250cm³ of solution G ( 2 marks)
        4. Calculate the:
           1. Molar mass of acid H₂MO₆ ( 3 marks)
           2. Mass of M in the formula H₂MO₆ given that (H = 1, O=16). ( 3 marks)
2. You are provided with solid A. Carry out the tests below. Identify any gas or gases produced and record your observations and inferences.
   1. Heat gently a spatula endful of solid A in a dry test tube. Test the gas with red and blue litmus paper.
      
   2. Place a spatula endful of solid A in a boiling tube. Add 10cm³ of distilled water. Filter off the residue. Divide the filtrate into two portions. Retain the residue.
      1. To the first portion add 3 drops of ammonia solution and then excess.
         
      2. To the second portion add 3 drops of Lead (ii) nitrate solution.
         
   3. To the residue obtained in (b) above add 5cm3 of dilute nitric acid. Divide the solution into two portions.
      
      1. To the first portion add 3 drops of Ammonia solution and then excess.
         
      2. To the second portion add 3 drops of lead (ii) nitrate solution.
         
         
         

Confidential

• Requirements for candidates

In addition to the apparatus and fittings found in a Chemistry laboratory, each candidate will require the following.

1. about 100cm³ of solution F
2. about 50cm³ of solution G
3. one burette 0 – 50ml
4. one pipette 25ml
5. two conical flasks
6. 100ml measuring cylinder
7. 200ml or 250ml beaker
8. label sticker
9. About 150ml distilled water
10. Phenolphthalein indicator
11. one CLEAN METALLIC spatula
12. one boiling tube
13. 6 clean dry test-tubes
14. one test-tube holder
15. blue and red litmus paper
16. filter paper

Access to:

1. Means of heating(Bunsen burner)
2. 2M ammonia solution with a dropper
3. 2M nitric(v) acid with a dropper.
4. 0.2M lead(ii)nitrate solution

NOTE

• Solid A is a mixture of Zinc carbonate and anhydrous zinc sulphate in the ratio 1:1.
• 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 23.46g of tartaric acid ( 2,3 dihydroxy butanedioc acid) in 200cm³ of distilled water and diluting it to 250cm³ solution.

Marking scheme

1.  
   1. Complete table ---------                                                                                                                                     1mk
      Must have 3 titrations done for       
      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
   2.  
      • Use of decimals – 1mk
        Tied to 1^st and 2^nd rows only
        1. Accept 1 or 2 dec. places used consistently
        2. If 2^nd 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 :
        1. If any titre is within ± 0.1 of s.v  1mk
        2. If none is within ±1 of s.v but least within ± 0.2 s.v award 1mk
        3. If none is within ±2 of s.v  0 mk
      • Principle of Averaging  1mk
        Condition:
        1. If 3 consistent values are averaged        1mk
        2. If 3 titrations done and only 2 are possible and averaged    1mk
        3. If any 2 titrations are done inconsistent and  averaged        0mk
        4. If 3 titrations are done, all are possible and only 2 averaged     0mk
        5. If 3 titrations are done are  inconsistent and  averaged     0mk
           Penalties
           1. Wrong Arithmetic i.e error outside ± 2 units in the 2^nd place penalise ½ mk
           2. If no work is shown but answer given is correct penalise ½ mk
           3. If the answer is rounded off to the 1^st place penalise ½ mk
           4. 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
        1. If within ±1 of s.v   …………………………..                                          1mk
        2. If within ±2 of s.v   …………………………..                                          ½ mk
        3. If beyond ± 0.2 of s.v …………………………..                                          0mk
           Summary
           CT - 1mk
           Dec –1mk
           AC-  1mk
           PA-  1mk
           FA-  1mk
                05
           
           CALCULATIONS
   3. 
      
      1. 25 x 0.1 = correct Ans
           1000
         Penalties
         1. Penalise fully for strange figure
         2. Penalise ½ mk for wrong answer if error is outside ± 2 units in the 4^th place
         3. Accept answer given to at least 4 dec. places otherwise penalise ½ mk
         4. Units may not be shown, but if shown MUST be correct otherwise penalise ½ mk for wrong units
      2. 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
      3. 100 x Answer C(II)        = correct answer
           Titre volume √½ mk              √½ mk
         Penalties
         1. Penalise ½ mk for WT (wrong transfer) of titre, otherwise penalise fully for strange figure
         2. Penalise ½ mk for wrong answer if the error is outside ± 2 units in the 4th dec place
         3. Treat as in (iii) – (iv) in C(i) abov
         4. 10cm³ diluted to 100cm³ therefore number of moles in 20cm³ is equal to moles in 100cm³  = correct answer √½
      4. Answer III same as IV
         Penalties
         1. penalise ½ mk for wrong Transfer (WT) otherwise fully for strange value
         2. Penalise ½ mk for rounding off answer to atleast 3 dec places
      5. Answer IV x 250         =   correct answer
                20  √½ mk                      1mk
         Penalties
         Treat as in (i) – (iv) in C I above
   4.  
      1. Molar mass =    23.46 √1mk
                               Answer V
                          = correct Answer √1mk
         Penalties
         1. penalise ½ mk for WT of answers in V, otherwise penalise fully for any strange figure used in the calculation
         2. Same conditions for units
         3. penalise ½ mk for not rounding off answer to a whole number
      2. H₂MO₆ = (2x1) + M+(6 x 16) = Answer dI
                                        √½    
                   = M + 98    = Answer d I                                             
                                              √½
              M = Answer d I – 98
              M = Correct answer √1
2.  
   1. 
      

      Observations	Inferences
      -Solid turns to yellow and then to white on cooling. (1 mk) -A gas that turns moist blue litmus paper to red.(1mk)	Zn2+ present (1 mk) The gas is acidic (1mk)

   2.  
      1. 
         

         Observations	Inferences
         A white precipitate soluble in excess. (2 mk)	Zn2+ present (1 mk)

          
      2. 
         

         Observations	Inferences
         A white ppt present (1 mk)	SO42-, Cl-, CO32-  present (2 mk for any two)

   3. 
      

      Observations	Inferences
      Effervescence present (1 mk)	CO32- present (1 mk)

      
      
      1. 
         

         Observations	Inferences
         A white ppt which dissolves in excess (2 mk)	Zn2+  present (1 mk)

      2. 
         

         Observations	Inferences
         No white ppt (1 mk)	SO42-, CL-, SO32- absent (2 mk for any two)