2018/2019 WAEC Chemistry Practicals Answers

2018/2019 WAEC Chemistry Practicals Answers

Question 1 
Consider the electron configuration of the following elements:
Q -1s22s22p1;
R - 1s22s22p63s2;
S - 1s22s22p63s23p1;
T - 1s22s22p6
U - 1s22s22p63s23p5.
Which element(s) Belong(s) to the same group?
                              Belong(s) to the same period?
                              Ionize(s) by readily:
                              Losing electrons;
                              Gaining electrons?
                              Is/are noble gas(es)?
                              Would react to form electrovalent compounds? [8 marks]

(i)   Define electronegativity.
State the trend of electronegativity on the periodic table.
Explain briefly why atomic radius increases down the group but decreases across the period in the periodic table. [7 marks]

List three characteristic properties of transition metals.                           [3 marks]
If 15 g of potassium trioxochlorate (V) is heated in the presence of MnO2,
write a balance equation for the reaction.
state the role of MnO2 in the reaction.
calculate the mass of chloride produced.
  [O = 16.0; Cl = 35.5; K = 39.0]                                                  [7 marks]
OBSERVATION: This questions was popular to the candidates and their performance was good. In (a), majority of the candidates were able to identify the elements that belongs to the same group and the same period. In (b), majority of the candidates did not give a complete definition of electronegativity while others defined it using element in place of atom in the definition.  Also candidates lost marks because they could not correctly state the trend of electronegativity in the periodic table and could not adequately explain why atomic number increase down the group but decrease across the period.
In (c), Majority of the candidates were able to list the properties of transition elements, though some candidate listed the general properties of metals.

In part (d), majority of the candidates were able to write and balance the equation for the thermal decomposition of kClO3 and were also able to state the role of MnO2 in the reaction.
The expected answers include:
I.(a)     (i) Q and S           
            (ii) R, S, U.           Q, T     
            (iii) I.   Q/R/S             
       II.  U                         
            (iv)       T                     
            (v)        R, U;  S, U;  Q, U;    

 (b)       (i)  Electronegativity is defined as the power of an atom to attract (shared pair or 
                   bonded) electrons in a molecule.     
            (ii) Electronegativity of elements increases across the period and decreases down 
                  the group.  
            (iii) There is an increase in the number of shells down the group and less nuclear
                    attraction hence increase in the atomic size.
Across the period, effective nuclear charge increases and the outermost electrons are drawn more towards the nucleus.   
Number of shells remains the same while nuclear charge increases.                                                           
(c) variable oxidation states
     formation of coloured ions
     complex ion formation
     paramagnetism/magnetic property
     catalytic ability                                                                                                                           
(d) (i) 2KClO3  →   2KCl + 3O    
      (ii) MnO2 acts as catalyst 
     (iii)       M(KClO3) = 39 + 35.5 + (3 x 16) 
                                     = 122.5 gmol-1      
                        M(KCl) =  39 + 35.5  
                                      = 74.5 gmol-1
                        (2 x 122.5g) of KClO3 gives (2 x 74.5g) of KCl
                        15g of KClO3     = 2 x 74.5 x 15    
                                                      2 x 122.5
                                                  =  9.12 g 

Question 2
(a) (i) State Boyle’s law.
Give the mathematical expression of Boyle’s law.
Give two reasons why real gases deviate from ideal gas behavior.
Three postulates of the kinetic theory of gases;
Dalton’s law of partial pressures
(b)    (i)      Consider the figure as illustrated below:
                                        P                             Q                        
                                      Gas                    Solid
                        State the processes represented by P, Q and R respectively
           (ii)    Some solids when heated change directly to the gaseous state.

What name is given to this phenomenon?
Name two substances which exhibit the phenomenon referred to in 2(b)(ii) I.
(c)        If 80 cm3 of methane (CH4) diffuses through a porous membrane in 20 seconds while 50   
cm3 of gas Q diffuse through the same membrane in 25 seconds, calculate the molecular mass of gas Q.
[ CH4 = 16.0 ]
Explain briefly why at room temperature fluorine is a gas while bromine is a liquid.
This question was attempted by majority of the candidates and the performance was fairly good.
In (a), majority of the candidates gave correct statement of Boyle’s law as well as the mathematical expression, but could not give reasons why real gases deviate from ideal gas behaviour.

In (b), processes of freezing, condensation and sublimation were easily identified by majority of the candidates.  In (c) some candidates could not solve the problem, however few candidates were able to solve the problem using Graham’s law of diffusion.
In (d) Some candidates could not explain correctly why bromine is a liquid and fluorine a gas at room temperature.
OBSERVATION: PThe expected answers include:
(a) (i) Boyle’s law states that the volume of a given mass of gas is inversely 
           proportional to its pressure provided temperature remains constant.
     (ii) V a  (1) at constant T. OR PV = K; OR V =  OR P1V1 = P2V2
    (iii) There are attractive forces between molecules of real gases.  
           The volume of molecules of real gases is not negligible.       
    (iv) I Motion of gas molecules is random or chaotic.
  1.  Collision between molecules is perfectly elastic/no energy is lost on collision.
  2.  The volume of the molecules is negligible compared with the total volume of the container.
  3.  There are no attractive or repulsive forces between molecules of the gas.
  4.  The average kinetic energy of the molecules is proportional to the (absolute) temperature.
  5.  The pressure exerted by the gas results from the impact/collision of molecules upon the walls of the container.                                                                                  
  II Dalton’ Law of partial pressures states that for a mixture of gases which do not react together the         total pressure is the sum of the partial pressures of the gases present.
            Accept mathematical expression PT = Pa + Pb + Pc where the terms are 
            Explained and candidates must state that gases do not react together                                                                                                                                                      
(b) (i) P  -  Condensation      
          Q  -  Freezing/Solidification   
          R  -  Sublimation    
    (ii) I.  Sublimation     
         II. Iodine, camphor, naphthalene, aluminum chloride            
(c)          =   Þ=   = 4          
              =   = 2
              =   =  = 
            4 =  Þ  = 4 x 16
                       = 64  
(d) Fluorine and bromine are both held by weak van der Waals forces.  
The van der Waals forces increase with increasing number of electrons.
Bromine having more electrons than fluorine has a stronger force which
holds the molecules closer together hence bromine is a liquid.
Fluorine having less number of electrons has weaker van der Waals forces 
hence the molecules are not tightly held together.

Question 3                                                                                                               
(a)        (i)     What is an acid-base indicator?
           (ii)     When is an indicator said to be most suitable for an acid-bas titration?
          (iii)     Consider the acid-base titration reaction represented by the following equation:
                                    NaOH(aq) + CH3COOH(aq)           CH3COONa(aq) + H2O(l)
            I.          Name a suitable indicator for the reaction.
            II.        Explain briefly your answer in 3(a)(ii) I.                                                                                                                      
(b) (i) Write the ionic equation for the reaction between zinc and silver trioxonitrate (V) 
    (ii) Which type of reaction occurs between zinc and silver trioxonitrate (V) solution?
           Give a reason for your answer.
(c) (i) Consider the reaction represented by the following equation:
           CaCO3(s) + 2HCl(aq)       CaCl2(aq) + CO2(g) + H2O(l).       
            State three factors that could increase the rate of the reaction.
 Explain briefly the observation that increase in temperature generally   
 increases the rate of reaction. 
(i)  What is solubility?
(ii) Distinguish between saturated solution and unsaturated solution.
A saturated solution of volume 10 cm3 yielded 0.06 g of its dry salt at 25oC.
 Calculate the solubility of the salt in gdm-3.
This question was fairly attempted by candidates.  In (a), candidates were unable to correctly define an acid-base indicator and were unable to explain the suitability of an indicator for a particular acid-base titration.  Some candidates used colour in acid and base as a bases for defining indicators instead of changes in pH.
In (b), a majority of the candidates could not write ionic equations although a few of them could identify the reaction as displacement/redox.
In (c), most candidates gave general factors instead of being specific e.g. “concentration” instead of “increase in concentration of HCl” “surface area.” instead of surface area of CaCO3”.
In (d), candidates did not give the correct/complete definition of solubility.  In defining solubility, some candidates lost marks for writing “in 1dm3 of solution” instead in 1 dm3 or solvent.
 The expected answers include:
(a) (i) An acid-base indicator is a weak acid or weak base (or organic compound) that has 
one colour in acid medium and another colour in alkaline medium.  
An acid-base indicator is a weak acid or weak base whose colour in the dissociated 
form is different from the colour in the undissociated form.                                                                      (ii)  An indicator is suitable, when it changes its colour (sharply) at the equivalence       
          point of the titration reaction. 
(iii) I. Phenolphthalein          
      II. CH3COONa formed in the reaction is hydrolysed in water to give 
           excess hydroxide ions.            
The resulting solution at equivalence point is alkaline and phenolphalein    changes colour in alkaline medium                                                                                                                     
(b) (i) Zn(s) +2Ag+(aq) → Zn2+(aq) + 2Ag(s)      
     (ii)  Redox reaction because there is a transfer of electrons from zinc to Ag+ .
Displacement because zinc displaces Ag+ ions from its aqueous solution                                              (c) (i) Temperature;                    
          Concentration of HCl     
          Surface area of CaCO3  
    (ii) Increase in temperature results in the particles gaining kinetic energy.  
           This causes the particles to move faster and collide more frequently.
           Effective collision is increased hence an increase in rate of reaction.              
(d) (i) Solubility is the maximum amount of the solute (moles or grams) that will
           saturate/dissolve in 1.0dm3 of solvent at a particular temperature.     
     (ii) At a particular temperature a saturated solution cannot dissolve any more 
            solute but an unsaturated solution can dissolve more solute.
     (iii)10 cm3 of saturated solution contains 0.06 g  
                        \1000 cm3 = 1000 x 0.06 g   
                          = 6.0 gdm-3  
Question 4
(i)  Define hybridization?
(ii) State the type(s) of hybridization exhibited by carbon in propene.
(iii) Mention the shape of the
Diamond and graphite are both allotropes of carbon.  Diamond is hard while graphite is soft.  Explain briefly this observation.              
A monochloroalkane was analysed by converting all the chlorine into chloride ions and precipitating the chloride as silver chloride.  If 0.08226g of the chloroalkane gave 0.1837g of silver chloride,
 calculate the molecular mass of the chloroakane;
(ii)  determine the molecular formula of the chloroalkane;
(iii) draw the structure of the chloroakane and name it.
                       [ H = 1.00; C = 12.0; Cl = 35.5; Ag = 108.0 ]
OBSERVATION:  This question was not so popular with the candidates.  Only few candidates attempted this question.  In (a), majority of the candidates do not have good knowledge of hybridization.  In (b) some candidates could not explain why diamond is hard and graphite soft.  In (c), a fair attempt was made at the calculation of the molecular mass of the choroalkane and the drawing of its structure.
The expected answers include:
(a) (i) Hybridization is defined as mixing of different atomic orbitals 
          to produce identical new orbitals (of the same shape and energy).                                                                                            OR
Hybridization is the mixing of two or more orbitals to give new sets of two or
more orbitals which are exactly equivalent.
(ii)    sp3  and sp2 
(iii)   I. s – orbital  -   spherical  
        II. p – orbital -   dumb-bell/pear shape/ figure eight
(b) In diamond each carbon atom is sp3 hybridized while in graphite each carbon atom is sp2 hybridized. Diamond has a giant covalent structure with a network of strong covalent bonds holding each atom tightly into crystals.

In graphite, each carbon atom is covalently linked to three neighbouring atoms 
in the same plane forming layers of carbon atoms held together by weak 
van der Waals forces.                                                                          
(c)        (i)         R – Cl     Ag+    AgCl    
                        1 mole             1 mole
                        M(AgCl) = 108 + 35.5 
    = 143.5gmol-1          
                        Moles of AgCl =  mass           
                                                     Molar mass 
                                                 = 1.280 x 10-3 moles   
I mole of RCl   = 1 mole of AgCl       
\mole of RCl  = 1.280 x 10-3 moles
                          Mass of RCl   = 0.0826 g
                          Molar mass     =    0.0826      
                                                      1.280 x 108  
                                                                         =  64.52 gmol-1      

R – Cl     Ag+    AgCl    
M(AgCl)  =  108 + 35.4  =  143.5 gmol -1 
Moles of AgCl   =    mass     =  0.1837   
                              Molar mass   143.5
                            = 1.280 x 10-3
     143.5 g of AgCl    ≡ 35.5 g of Cl
     0.1837 g of AgCl ≡   35.5  x 0.1837     
                                  =  0.0454 g
  Hence 0.0454 g of Cl  = 0.0826 g of RCl
                 35.5 g of Cl  = 0.0826  x 35.5  
                                    =  64.52 gmol-1     

 R – Cl     Ag+    AgCl
 1 mole              1 mole
 M(AgCl)  =  108 + 35.5    =   143 . 5
 0.1837 g of AgCl produced from 0.0826 g of RCl  
  ...         143.5 g of AgCl will be produced from:
                                 0.082   x 143.5   
                                 =    64.5 gmol-1                  
(ii)        Since the substance is an alkyl chloride/ chloroalkane
                    CnH2n+1Cl = 64.5       
                      =  12n +2n + 1 +35.5 = 64.5
                       14n = 64.5 – 36.5
                        = 28                 
                     n =  = 2              
                     Molecular formulae = C2H5Cl
                                    H         H                                                            
(iii)       H         C         C         Cl       

                                    H         H


Question 7
(a) (i)Sodium hydroxide can be prepared by the electrolysis of concentrated solution of                                   sodium chloride.  Write the equation at the:
                        I.  cathode;
                       II.  anode.
     (ii) Explain briefly why the electrolysis of molten alumina is considered environmentally friendly              while that of molten sodium chloride is not.
(b) (i) Describe briefly the electrolysis of copper (II) tretraoxosulphate (VI) solution using                               copper electrodes.
     (ii) State the colour of the electrolyte in 7(b)(i):
             before the electrolysis;
             after the electrolysis.
     (iii) Give a reason for your answer in 7(b)(ii).
 (c) (i)  Name the impurities present in bauxite.
     (ii)  State how the impurities in bauxite can be removed..
(d) An aluminium of mass 3.14g reacted with hydrochloric acid at s.t.p.
      (i)    Write a balanced equation for the reaction
      (ii)  Calculate the:  mass of hydrogen produced;
              volume of hudrogen produced at s.t.p.
            [ H = 1.00; Al = 27.0; Molar volume = 22.4 dm3 ]
OBSERVATION:  This question was attempted by only few candidates and their performance was below average.  In (a), candidates made a fair attempt to write the equation at the cathode and anode but could not satisfactorily explain why the electrolysis of molten alumina is considered environmentally friendly while that of sodium chloride is not.
In (b), a fair attempt was made in describing electrolysis or CuSO4(aq) using copper electrodes.  some candidates were able to state the colour of the electrolyte before end after the electrolysis but could not give a correct reason for the answer.
In (c), majority of the candidates could neither name the impurities present in bauxite nor state how they are removed.
In (d), some candidates were able to write and balance the equation for the reaction between aluminum and excess hydrochloric acid at s.t.p. although many were not able to calculate the mass and the volume of hydrogen produced at s.t.p.
The expected answers include:
7. (a) (i) I.  Cathode:    Na+ + e- → Na (mercury cathode)/
                 2H+ + 2e-           H2 (Graphite cathode)      
            II.  Anode:   2Cl- → Cl2 + 2e-            

        (ii) During the electrolysis of molten alumina oxygen gas is produced.  
              During the electrolysis of molten sodium chloride chlorine gas is produced.     
              Oxygen gas is not a pollutant but chlorine gas is a pollutant.
(b)(i) At the cathode Cu2+ ions are discharged  preferentially and deposited as metallic copper on the            cathode.
At the anode, no ions is discharged but the conversion of copper atoms to ions is favored because it required less energy.  
                    Cathode                                 Anode
           Cu2+(aq)+ 2e-                   Cu(s)              Cu(s)              Cu2+(aq) +  2e-     
           Preferential discharge                    Copper atoms go into solution because it 
                                                                  requires less energy   
(ii)        I.          blue     
            II.        blue    
(iii)  No colour change in the electrolyte because the copper (II) ions discharged at the cathode are replaced by the dissolving copper anode.                                          
(c)  (i) Iron (III) oxides and trioxosilicates  (IV).  
      (ii) The impurities in bauxite can be removed by heating the bauxite with concentrated sodium hydroxide (under high pressure) to form (soluble) sodium aluminate  where the undissolved impurities are filtered off.                                                                                                         
(d) (i)  2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2 (g)          
     (ii) I.  mole (Al) = n =         
     = 0.1163mol
       But  = 
     = n(H2) =  
     = n (H2) = 0.174mol    
    = Molar mass H2 = 2 x 1 =2
    =M (H2) = 2 x 0.174
   = 0.3489(g) = 0.349g               
     II.1 mole of a gas at s.t.p = 22.4 dm3     
                \ 0.174 mole of H2 gas = 0.174 x 22.4
                 3.8976 dm3= 3.90 dm3                                                                                                                                
 (d) (i)  2Al(s)   +  6 HCl(aq) 2AlCl3(aq)  + 3H2(g) 

     (ii) I.2 x 27 g of Al requires 3 x 2 g of H2
             54 g of Al produce 6 g of H2   
              3.14 g of Al     =   6   x 3.14 g    
                                =  3.49 g 
II.  2 g of H2         ≡ 22.4 dm3  
      0.349 g of H2  ≡ 22.4  x 0.349
                                    =   3. 91 dm3 

 Question 8
Potassium , hydrogen gas and potassium hydride exhibit different types of bonds.
Copy and complete the following table

Substance    Type of Bonding        Units which make-up this substance

       K                     H2                                               KH

(i) Arrange the following compounds in order of increasing boiling points:
      H2O, CH4, H2S. 
(ii) Give reasons for your answer in 8(b)(i).
(c) (i) State two processes where the concept of oxidation number is applied.
     (ii) Write the formula of the oxide in which nitrogen exhibits an oxidation number of:

(iii) Differentiate between a drying agent and a dehydrating agent.
       Give an example of each agent.
(d) (i) State two chemical properties of an acid.
      (ii) Classify each of the following solutions as acidic, basic or neutral;

A fair attempt was made on this question.  Majority of the candidates could not answer (a) part of the question especially the unit which make up the substance.
In (b), only few candidates could arrange the compounds in order of increasing boiling point.
In (c), candidates gave a fair attempt of application of oxidation number and also wrote formulae of compounds in which nitrogen exhibits oxidation number of +1, +2 and +3.  Many candidates could not effectively differentiate between a drying agent and a dehydrating agent.
In (d), majority of the candidates that attempted the question correctly gave properties of an acid and they were also able to classify the following solutions NH4 C(aq), K2 CO3(aq);
Na2SO4(aq) and CH3COONa(aq) as either acidic, basic or neutral.
The expected answers include:

Substance  Type of Bonding    Units which make-up this substance

    K                         Metallic                K+ surrounded by mobile electrons 
    H2                       Covalent               H2 molecules   
    KH                      Ionic                      K+ ions  H- ions                                                                                                                          
(b) (i)                         
             Increasing boiling points
    (ii) CH4 – weak van der Waal’s forces     
          H2S - Strong van der Waal’s forces than CH4                        
          H2O – Hydrogen bond which is stronger than van der Waal’s forces    
(c) (i) Oxidation number is applied in the IUPAC system of naming 
Balancing redox equations.
Oxidation number identified which element in a compound is oxidized or            
 reduced /redox reactions.                                       
   (ii) I.  N2O     
        II.  NO      
      III.       N2O3     
(iii) Drying agent removes molecules of water or moisture from substances while dehydrating agent removes elements of water from compounds. 
Drying agent removes water from a substance without affecting the chemical composition while a dehydrating agent removes element of water from a substance and affects the chemical composition 
            Drying agent – Quicklime/Calcium chloride/Copper (II) oxide/Silica gel
            Dehydrating agent – Conc. tetraoxosulphate (VI) acid
(d)  (i) Reacts with metals to liberate hydrogen.
            Reacts with bases to form salts and water (only).
            Reacts with trioxocarbonate (IV) to liberate carbon (IV) oxide.
  (ii) I. NH4Cl(aq)- Acidic        
       II. K2CO3(aq)  -  Basic       
      III. Na2SO4(aq) – Neutral 
     IV.  CH3COONa – Basic 
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