Chemistry Paper 3 Questions and Answers - Form 3 Term 2 Opener Exams 2023

Instruction to candidate.

• Attempt all the questions

The paper has a maximum score of 30 marks.

You are provided with:-

• Solution A, Hydrochloric acid.
• Solution B, 0.024 M Sodium hydroxide.
• Solution C, containing 15.74g of Na₂CO₃ . xH₂0 in 250ml of the solution.

You are required to:-

1. Prepare a dilute solution of the hydrated sodium carbonate, C.
2. Determine:-
   1. The concentration of solution A.
   2. The value of x in the carbonate.

Procedure a

• Using a pipette place 25.0 cm³ of solution C into a 250ml volumetric flask.
• Add about 200cm³ of distilled water. Shake well.
• Add more distilled water to make upto the mark.
• Label this solution D.
• Retain solution D for use in procedure b and c.

Procedure b

• Fill a burette with solution A.
• Using a clean pipette and pipette filler, place 25.0 cm³ of solution B into a 250ml conical flask.
• Add two drops of phenolphthalein indicator and titrate with solution A.
• Record your results in table 1.
• Repeat the titration two more times and complete the table.

Table 1

(4 marks)

1. Determine the:-
   1. Average volume of solution A used. (show your working) (2marks)
   2. Number of moles of sodium hydroxide in 25cm 3 of solution B used. (1 mark)
   3. Number of moles of acid in volume of solution A used. (2marks)
   4. Concentration of solution A in moles per litre. (2marks)

Procedure c

• Fill the burette with solution A. Using a pipette, pipette 25.0cm 3 of solution D into a conical flask. Add 2 drops of methyl orange indicator and titrate with solution A.
• Record your results in the table.
• Repeat the titration two more times and complete the table.

Table 2

 (4 marks)

b. 

1. Determine the:-
   1. Average volume of solution A used. (2marks)
   2. Moles of the acid in the average volume of solution A used. (2marks)
   3. Concentration in grams per litre of the carbonate in solution C. (2marks)
2. Write an equation for the reaction that occurred between the acid and the carbonate (1mark)
3. Determine:-
   1. number of moles of the carbonate in 25cm³ of solution D used. (2 marks)
   2. Number of moles of carbonate in 250cm³ of solution D. (2 marks)
   3. Concentration of solution C in moles per litre. (2marks)
   4. Value of x in Na₂CO₃.xH₂O.   (H= 1.0, C = 12.0, O = 16.0 Na = 23.0) (2 marks)

CONFIDENTIAL

Ensure each student have the following:

1. 150cm³ of solution A labelled A.
2. 100cm³ of solution B labelled B.
3. 3 conical flasks.
4. 50ml burette.
5. 25ml pipette.
6. One complete retort stand.
7. 250ml volumetric flask.
8. A white tile.
9. 1 label
10. 500ml distilled water in a wash bottle.
11. 250ml empty beaker.
12. Methyl orange indicator in a bottle dropper. (freshly prepared)
13. One filter funnel
14. 50cm³ of solution C labelled C
15. Phenolphthalein indicator in a bottle dropper

N/B

1. Solution A is made by dissolving 4.3cm³ of concentrated hydrochloric acid (1.18g/cm³ ) in 600cm³ distilled water and diluting to 1 litre.
2. Solution B is made by dissolving 0.96g of Sodium Hydroxide pellets in about 800cm³ of distilled water and diluting to 1 litre.
3. Solution C is made by dissolving 15.74g of hydrated Sodium Carbonate – Na₂CO₃.10H₂O – in 800cm³ of distilled water and diluting to 1 litre solution.

MARKING SCHEME

Procedure (b)

Table 4 mks    –

• Complete table – 1 mk
• Use of dec – 1 mk
• Accuracy:
  c in 0.1 cm³ of S.V – 1 mk
  c in 0.2 cm³ of S.V– ½ mk

Rej.    

• Unrealistic values i.e. beyond 50cm³.
• Wrong arithmetic

1.  
   1. Average volume of solution A used.
      constant + Values ✓½ = _______________Ans ✓½
                   No.
      12.1 + 12.0 + 11.9 = 12.0 cm^3­           ( 1mk)
                    3     
   2. No. of moles of NaOH in 25cm³ of solution B used = M = ⁿ/_v = n = MV
      = (0.024 × ²⁵/₁₀₀₀) ✓½ = 0.006 ✓½ mole
   3. Moles of acid in volume of Soln A used.
      NaOH_(aq) + HCl_(aq) →  NaCl_(aq) + H₂O_(l)
      Mole ratio: NaOH : HCl  is 1 : 1
      Therefore moles of acid = Ans. In (II) above ✓½
      i.e. 0.0006 mole         ( 1 mk)
   4. Concentration of soln A moles/litre
      n =     Ans (III) above        or Ans.III✓½ × 1000 = Ans ✓½
      v      Ans I above ÷ 1000                Ans I
      0.0006 x 1000= 0.05 moles/l
       12.0
      Procedure C
      Table 2: As for table 1. Total (4 mks)
2.  
   1.  
      1. Average volume of solution  A used.
         Constant values✓½   = cm³✓½
                    No.                    
          e.g 22.1 + 22.0 + 21.9 = 22.0 cm³  1mk
                            3
      2. Moles of acid in the average volume of solution A used.
         V =  MV       Ans. A(iv) above x ans bi (I) above
               1000                                           1000             = ___ cm³
         e.g 0.005 x 22.0 x ¹/₁₀₀₀ = 0.0011 moles (1 mark)
      3. Concentration in grams /litre of carbonate.
         Mass (g)  = 15.7   or  15.7 x 1000✓ ½
           = 62.9 g/l ✓ ½
   2. Equation
      NaCO₃. xH₂O _(aq)®CO₂ (g) + 2NaCL(aq) + (x + 1) H₂O (l)        (1mk)
   3.  
      1. No. of moles of carbonate in 25cm³ of solution D
         Mole ratio: Carbonate: Acid 1:2
         Therefore moles of carbonate used = ½ × Ans ✓½ (bi, II) above
                                                                  =____________ ans ✓½
         e.g  ½ x 0.0011 ✓ ½  = 0.0055✓ ½  moles
         ANS =  0.00055 mol
      2. Moles of carbonate in 250cm³ of solution D.
         = Ans. B iv, I above x ²⁵⁰/₂₅ = ______ans
         e.g 0.00055 x 250/25 = 0.0055 moles
      3. Concentration of solution c ( moles /litre)
         0.0055 moles = moles in 250ml of D = moles in 25 ml of C
         Molarity = u/v
         ( ans in b iv,ii above x ¹⁰⁰⁰/₂₅) ½ mk = _______ans ½ mk            
         e.g 0.0055 x ¹⁰⁰⁰/₂₅ = 0.22 ½ moles /l
      4. value of x in Na₂CO₃.xH₂0
         Molar mass = 106 +18x
         Mass used = ans III above (106 + 18x) = ans (ii) above
                                                                       = ans_____√ ½
         e.g 0.22( 106 + 18x) = 62.9 √ ½
         x =       39.64. 
                       3.96
         = 10.01
         ≈ 10 √ ½