waecE-LEARNING
Chemistry Paper 3,Nov/Dec 2012  
Questions:   1 2 3 Main
General Comments
Weakness/Remedies
Strength

























































Question 1

                                                                                                                                                    

IQuestion 1

  1. The following table shows burette reading obtained when 25.0 cm3 of a solution of sodium trioxocarbonate (IV) containing 1.30 g in 250 cm3 of solution was titrated against a solution of hydrochloric acid of an unknown concentration:

 

Burrette reading

Rough

I

II

Final reading (cm3)

23.50

44.20

25.20

Initial reading (cm3)

2.00

23.50

4.30

Volume of acid used (cm3)

 

 

 

  1. Copy and complete the table.
  2. Calculate the average volume of acid used.
  3. Write a balanced equation for the reaction in the titration

 

  1. From the information obtained in 1(a) above, calculate the:
  2. concentration of Na2CO3 in moldm‑3;
  3. concentration of HCl in
  4. moldm-3 ;
  5. gdm-3.

[ H = 1.00; O = 16.0; Na = 23.0; Cl = 35.5 ]

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OBSERVATION

TThis question was attempted by majority of the candidates and the performance was good.
In (a)(i) candidates were required to copy and complete a table of titre values.  Majority of the candidates performed well, but a few of them could not still subtract the initial from final readings.
In (a)(i) and (ii), candidates correctly gave the volumes of B used and went on to use consistent titre values in averaging. However, a few candidates lost marks for using non-concordant titres.
in (a)(iii), majority or the candidates correctly balanced the chemical equation.  Some candidates lost marks as a result of writing wrong physical state.  Candidates should note that balanced chemical equation would be marked even without physical state, but writing the wrong physical state will result to forfeiture of some marks.
In (b), most candidates performed very well in this question but quite a number of them lost marks for wrong mole ratio and for not correctly giving their answers to 3 significant figures.

 

 

 

 

 

The expected answers include:

1.(a)(i)

 

Burette reading

 

Rough

 

I

 

II

 

Final reading (cm3)

 

23.50

 

44.20

 

25.20

 

Initial reading (cm3)

 

2.00

 

23.50

 

4.30

 

Volume of B used (cm3)

 

21.50

 

20.70

               
20.90

             
(ii)  Average titre value = 20.70 + 20.90        
                                                                   2
                                                       = 20.80 cm3                                                            

(iii)  Na2CO3 + 2HCl → 2NaCl + H2O + CO2                      
              
                                      
(b)(i)       250 cm3 contains 1.30 g of Na2CO3
\  1000 cm3 will contain 1.30 x 1000           
         250
                                                  = 5.20 gdm-3 
               
Molar mass of Na2CO3 = (23 x 2) + 12 +(16 x 3)      
= 106 gmol-1                    

            \ Concentration in moldm-3   =   5.20                      
                                                                  106
                                                    = 0.0491 moldm-3                                                                                                           
                                                                                               
(ii) I                CaVa = 2                                
                        CbVb   1                                                  

                        Ca x 20.80   =  2
                        0.0491 x 25      1        

            Ca   =   2 x 0.0491 x 25         
                            20.80
           
                   =   0.118 moldm-3            
                                                                                
      II.  Concentration in gdm-3  ρ = Conc. x Molar mass
            Molar mass of HCl = 1 + 35.5 = 36.5gmol-1                            
            Conc. gdm-3 = 0.118 x 36.5               
                                    = 4.31gdm-3                                 
           

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