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Chemistry Paper 2,Nov/Dec 2008  
Questions:   1 2 3 4 5 6   Main
General Comments

Question 5

(a)        (i)         State how the following procedures would be carried out in the laboratory:
                        I.          soften temporary hard water without heating it;
                        II.         obtain pure water from muddy water;
                        III.       remove sediments in a sample of water.

            (ii)        State the function of
                         I.        alum,
                        II.         chlorine in a water treatment plant.

            (iii)       State two advantages of soft water over hard water.                      [8 marks]

(b)        (i)         Name one metallic oxide which increases in mass without dissolving when 
                        exposed to the atmosphere.

            (ii)        What phenomenon is exhibited in 7(b)(i) above;

            (iii)       Write a balanced chemical equation between the metallic oxide in 7(b)(i)
                         above and dilute trioxonitrate (V) acid.                                          [4 marks]

(c) Consider the following diagram for the laboratory preparation of a gas W.








            (i)         Identify W, X, Y  and Z.
            (ii)        State the function of the manganese (IV) oxide.
            (iii)       Write a balanced chemical equation for the preparation of the gas W.
            (iv)       Give the reason why W cannot be collected by the displacement of air.
                                                                                                                                [8 marks]

(d)        Calculate the number of molecules in 2.8g of nitrogen gas.
                           [ N = 14, Avogadro’s constant = 6.02 x 1023 mol-1]                    [5 marks]


This question was attempted by very few  candidates and the performance was fair.
In(a)(i), candidates correctly stated the procedures that would be carried out in the laboratory for each of I – III as follows:

            I.          Add washing soda (Na2CO3 10H2O)/pass it through an ion-exchange
                        resin add calculated amount of slaked lime (Ca(OH)2).
           II.          Distillation
          III.         Filtration (accept decantation)/centrifuging.

In (a) (ii), candidates stated correctly that alum is for coagulation of fine/colloidal particles and chlorine is for killing bacteria in a water treatment plant.


In(a)(iii), candidates correctly gave two advantages of soft water over hard water from among the following:

  1. it does not waste soap/lathers readily with soap
  2. it does not cause furring/scales/deposits in kettles and boilers
  3. it is cheaper when used in dyeing and tanning industries.


In (b)(i), candidates gave the name of a metallic oxide which decreases in mass without dissolving in it when exposed to the atmosphere as calcium oxide/copper(II) oxide/magnesium oxide.

In (b)(ii), most of the candidates knew that the phenomenon is hygroscopy.

In (b)(iii), candidates could not write a balanced chemical equation for the reaction between the metallic oxide and dilute trioxonitrate (V) acid.  The expected answer was

                        CaO(s) +  2HNO3(aq),  →   Ca(NO3)2(aq) + H2O 
                        CuO(s) +  2HNO3(aq),  →   Cu(NO3)2(aq) + H2O 

In (c)(i), candidates could not identify W, X, Y and Z.

The expected answers were:

W        -           oxygen/ O2(g)
X         -           flat bottom flask
Y          -           conc  H2SO(4)/fused   CaCl2/U-tube
Z          -           mercury

In (c)(ii), candidates knew that manganese (IV) oxide is acting as a catalyst.

In (c)(iii), candidates could not give a balanced chemical equation for the preparation of the gas W
   2H2O2(aq)  ii    2H2O(l) +  O2(g)

In (c)(iv), candidates knew that W cannot be collected by the displacement of air because it has nearly the same density with air.                                                                                                                                  

In (d), most candidates did not know that nitrogen is diatomic hence lost marks for using N = 14 instead of N2  =  28.  The expected solution to the question is as follows:

Number of moles of nitrogen  =  mass                   =   2.8
                                                     molar mass               28
                                                                           =   0.10 moles

            1 mole of nitrogen  =   6. 02 x 1023  moles
            0.1 moles of nitrogen will contain 6.02 x1023  x  0.10
                                            =  6.02 x 1022   particles

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