waecE-LEARNING
Chemistry Paper 2 (Essay) ,May/June 2012  
Questions:   1 2 3 4 5 6 7 8   Main
General Comments
Weakness/Remedies
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Question 2

  1. (a)   (i)   What is diffusion?
            (ii)   State Charles’ law.
           (iii)    Sketch a graph to illustrate Charles’ law.
           (iv)    A given mass of a gas occupied 150 cm3 at 27 0C  and a pressure of 1.013 x 105 Nm-2.
                     Calculate the temperature at which its volume will be double at the same pressure.
           (v)     Arrange the three states of matter in order of increasing:
                    (i)    kinetic energy;
                   (ii)    forces of cohesion.                                                                                                 [11 marks]  

    (b)   (i)  State Le Chatelier’s principle.
           (ii)  A metal M forms two oxides containing 11.1% and 20.0% of oxygen.
                  Show that these figures agree with the law of multiple proportions.                    [7marks]

    (c)    The table below shows the physical properties of substances A,B and C.


    Substance

    Melting point/oC

    Boiling point/oC

    Solubility in water at 25oC

    A

    30

    117

    Insoluble

    B

    31

    160

    Insoluble

    C

    861

    1200

    Soluble

    1. If Aand B are miscible when melted and B and C react when heated, describe how a mixture of A, B and C could be separated.
    2. When 25.25g of the mixture A, B and C was separated, 7.52 g of A and 8.48 g of B were recovered.  Assuming that there was no loss of components during the separation, calculate the percentage by mass of C in the mixture.                 [7 marks]                                                                      
         
                        

 

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OBSERVATION

                   This question was attempted by majority of the candidates and the performance was good.
          In (a)(i) and (ii); most candidates correctly define diffusion, state Charles’ law and draw the
          graph of Charles’ law. However some candidates did not correctly labelled their axes and
          some candidates left out the word absolute in their definition of Charles’ law. 
 

          In (vi) most candidates were able to calculate the temperature correctly.  In (v) most 
         candidates were able to arrange the three states of matter in order of increasing kinetic
         energy and forces of  cohesion.

 

          In b(I), most candidates correctly stated Le Chatelier’s principle but only a few candidates
          could correctly show by calculation that the figures agree with the law of multiple
         proportion.

 

          In (c) most candidates could not describe how mixture of A,B and C could be separated but
         correctly calculated the percentage by mass of C in the mixture.

         The expected answers were:
         
          (a)       (i)   Diffusion is the movement of particles from the region of  higher 
   concentration to the region of lower concentration.                         

 

 

 

         (ii)   Charles’ law states that the volume of a given mass of gas is directly
                  proportional to its absolute temperature at a constant pressure.                 
                         Accept  = K at constant pressure with each term defined.
                         OR

                          V µ T at constant pressure with each term defined.


(iv)       Applying Charles’s equation

            V1        =          V2                   T2 = T1V2
            T1                    T2                                V1

                        T2     =     300 x 300       
                                                150        =   600K    

            (v)    (i)        Solid  liquid  gas /Solid, liquid, gas.
                    (ii)       Gas   liquid   solid/Gas, liquid, solid.           
                         
(b)      (i)     In a reversible reaction at equilibrium, if any of the factors affecting the                    equilibrium is altered, the position of equilibrium will change so as to annul                    the effect of the change.

(iii)    Oxide II

         % of O2 = 20%
         % by mass of metal   = 80%
         20 of O2 combined      80 of metal                    \ 100 of O2             =  100 x 80        
                                          20 
                         =  400 g             
                                                         =  400.9               

                                                                                                                                      
          (ii)    Oxide I

                   % of O2 = 11.1 %
                    % by mass of metal   = 88. 9 %
                   11.1 of O2 combined 88.9 of metal 
                    \ 100 of O2    = 100 x 89 .9   = 800.9 g  
                                                 11 .1

        
               Ratio of metal in both oxides   = 800.9:  400 = 2 : 1                                       [7 marks]   

 

                                                                                                                                                                                                                    
(ii)        ALTERNATIVE
            Consider 100g of each oxide:
                                                                                     Oxide I                           Oxide II
            Mass of O2…………………..                         11.1g                              20.0g          
            Mass of metal ……………….                         100 – 11.1   = 88.9        100 – 20.0 = 80.0
            Mass of metal combined with 11.1g of O2      88.9                                80.0  x 11.1=44.4
                                                                                                                            20.0
            Ratio of masses of metal =   88.9 : 44.4                                                         
                                                     =   2:1

(ii)       ALTERNATIVE

            For sample A 11.1 g of O2 combined with 88.9 g of metal
            For sample B 20 g of O2 combined with 80 g of metal
                                                \ 11.1g O2 →   x    
                                                 =  44.4 g   
                                                 \ Ratio of metal I – A and B = 88.9 : 44.1
                                                 =    2 : 1    

             Note:   The mass of any of the elements can be fixed.

(c)        (i)                     -           Add water to mixture of A, B and C and stir; 
-           C dissolves;
-           Filter to obtain C as filtrate and A and B as residue;
-           Evaporate filtrate to dryness to obtain solid C;
-           Heat residue until melted;
-           Separate by simple distillation;
-           Pure A is collected at 1170C and pure B at 160oC.

            (ii)          Mass of mixture          =   25.25 g
                          Mass A+ B              =   7.52 g + 8.48 g
                                                             =    16.00 g

  • Mass  of  C            =     25.25  - 16.00 g

                        =     9.25 g       

  • % of C in mixture =     9.25  x 100

             25.25      
      =    36.6 %             

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