Chemistry May/June 2010

In part (a)(i), candidates correctly gave name and corresponding nature of the radiations emitted during radioactivity as follows:

In(a)(iii), most candidates correctly balanced the nuclear reactions and identify X and Y thus:

     (iii)               ²¹²₈₄PO → ²⁰⁸₈₂Pb +⁴₂ X                    X is α- particle/helium nucleus I⁴₂He

In part (b) (i) – (v), majority of the candidates correctly answered as follows: (i)            atomic number – 17 (ii)           group 7 (iii)           – formation of diatomic molecules – formation of monovalent ions/same valence electrons – they are coloured substances – they are oxidizing agents – most electronegative in its period/smallest atomic size in its period – they are non metals (iv)         chlorine (formula was not accepted) (v)          3Cl₂ + 6NaOH   →   5NaCl + NaCl0₃ + 3H₂0 However, some of the candidates lost marks because they could not give a correct balanced equation for the reaction of element X and hot concentrated NaOH.

Also the gave the name of the reaction as combination/neutralization in (iii).

ANSWER

In part(a)(i)saturated solution is that which contains the maximum amount of solute it can dissolve at a given temperature (in the presence of undissolved solute).

In part(a)(ii), only very few of the candidates could determine whether the solution formed was saturated or not as follpws:

Solubility of KN03 in in g dm-³ = 3.00 x 101 = 303
.. 1000cm³ of saturated solution = 303g
250cm³ of the solution = 303 x 250
                                                1000
                                           = 75.8 g

Since the quantity of KN03 added (67.0) to 250 cnr’ of water is less than maximum amount required to form a saturated solution, then the solution is unsaturatedj Any other correct method was accepted]

In part(b )(i), only few candidates knew that in dative bond, only one of the participating atoms/species donated electrons to be shared by both atoms while in covalent bond both participating atoms/species contribute equally the electrons being shared.

In (b )(ii), candidates could not explain why sugar and common salt do not conduct electricity in the solid state. The expected answer from candidate was that sugar is covalent while common salt (NaCl) is electrovalent/ ionic. Electrical conductivity (in compounds) depends on presence of mobile ions. Sugar does not conduct electricity because it does not contain ions (they are molecules) while solid common salt does not conduct electricity because its ions are not mobile.

In part (b )(iii), candidates knew that the intermolecular forces present in hydrogen fluoride and argon were hydrogen bond and van der Waal’s forces respectively.

In part (b)(iv), candidates knew that 0 – H —- N has the stronger hydrogen bond because oxygen is more electronegative and smaller in size than sulphur.

In part (c)(i) and (ii), most candidates could state Dalton’s Law of Partial Pressure but only very few of them could correctly determine the volume of the dry gas at s.t.p

The expected answers from candidates in(i) and (ii) respectively were:

(I)     the total pressure exerted by a mixture of gases which do not react chemically is equal to the sum of the individual partial pressures of the gases in the mixture
(II)    pressure of the dry gas (P 1) = 700 – 15 = 685 mmHg
VI = 200cm³, TI = 18°C = 273 + 18 = 291K, P2 = 760 mmHg,
T2 = 273
P1V1 = P2V2
  T1         T2
V2 = P1V1T2   =       
          P2T1

= 685 x 200 x 273
   760 x 291

= 169.1cm³ (169cm³ was accepted)

ANSWER

In(a)(i) and (ii), candidates correctly defined Arrhenius base as a substance which produces hydroxyl (OH-) ions in its aqueous solutions and gave one example from among NaOH, Ca(OH)2, KOH and NH40H (names were also accepted for the examples).

In(a)(iii),only few candidates correctly identified aqueous solutions of:
       I.              C_I2H₂₂0₁₁. as non – electrolyte;
       II.           NH3 as weak electrolyte;
       III.         NaOH as strong electrolyte.

In(a)(iv), most candidates could not write a balanced equation to represent the reaction between
CH3COOH and KOH. The expected answer from candidate was as follows.

In part(b), candidates correctly calculated the volume ofHCl required for complete nuetrilization of the NaOH thus:

CAVA          = 1
CBVB

VA = CBVB =
           CA               0.30 X 20
                               0.50             = 12.00cm3 (other method was accepted)

In parte c )(i) and (ii), candidates correctly gave the IUP AC name of each salt as sodium oxochlorate (1) and magnesium hydrogen trioxocarbonate (IV) respectively.

In parte d)(i), candidates correctly defined standard solution as a solution whose concentration is (accurately) known. However, most of them could not state the compound which is suitable or the preparation of a standard alkaline solution nor give reason for their answer in (d)(ii).

The expected response from candidates was Na2C03, because it is stable in air/non deliquescent.

In(d)(iii), candidates correctly wrote an equation for the reaction between Fe and dilute HCl in I and calculated the relative atomic mass of the metal in the given reaction in II as follows:

1. Fe(s) + 2HCl (aq) –+ FeCl2(aq) + H2(g)
II. Number of moles of HCI used = 2.20 x 50.0
                                                                      1000         = 0.11

Mole ratio ofHCl: Fe = 2: 1
:. number of moles of Fe used = 0.11 = 0.055 mole
                                                            2
0.055 mole of Fe = 3.08g
:. 1.0 mole of Fe = 3.08 x 1.0
                                     0.055        = 56g
       Relative atomic mass of Fe  = 56 (other method was accepted)

ANSWER

n part(a)(i), candidates gave two reasons why soda lime was used instead of caustic soda in the preparation of methane thus:

– soda lime does not attack glass apparatus unlike caustic soda
– soda lime is not deliquescent unlike caustic soda

In (a)(ii), candidates listed two physical properties of methane from among the following:
– it is a gas at room temperature;
– colourless gas;
– odourless gas;
– slightly soluble in water;
– less dense than air.

In (a)(iii), candidates determined correctly both empirical and molecule formulae of the hydrocarbon thus:

1. Empirical formula C2H5
II. Molecular formula (C2 H5)n = V.D x 2

→(12 x 2 + 1 x 5)n = 58
                          29n    =    58
                       n = 2
 →(C2 H5)2            = C4 H10

In part (b), candidates stated what was meant by isomerism, drew the structures of the two isomers of the compound C2H60, gave the name of the two isomers in (i) – (v) respectively.

However, some of them lost marks because of spelling errors, omission of bonds etc.
The expected responses from candidates were as follows:

  (i)            The existence of two or more compounds with the same molecular formula but
different molecular structures/different arrangements of atoms.

  (ii)     H    H
             I     I
        H-   C-   C-   O  -H
             I      I
           H       H

           H              H
             I               I
        H-   C-   O-   C  -H
             I                I
           H                 H

[All bonds must be shown however, accept -OH was accepted]
  (iii)           Ethanol and methoxymethane respectively.
  (iv)         The two isomers belong to different homologous series/they have different functional groups

In part (c), most candidates could not give three deductions made from qualitative and quantitative analysis of a given organic compound hence lost marks. The required answers from candidates were:

– functional groups
– number of atoms of different elements
– types of elements
– types of bonds
– percentage composition of the elements in compound
– spatial arrangements of atoms in a molecule.

In parts( d), only very few candidates could correctly give one chemical test to distinguish between propene and propane. The expected answer from candidates was as written below:

– Pass each of the gases into (acidified) KMnO4J bromine water/bromine in CCl4/ bromine, propene decolourizes bromine water/KMnO4/bromine in CCl4/bromine whereas propane does not.

ANSWER

In (a), candidates correctly named one product of destructive distillation of coal that was solid, liquid and gas in (i)-Ciii) as coke, arnrnoniacalliquour/coal tar and coal gas respectively.

In part (b) only few candidates knew that methane was the maj or component of synthetic gas and that synthetic gas was not a major source of air pollution because it does not contain sulphur or sulphur compounds/sulphur was removed in the gasification process.

In part (c), most candidates were able to correctly write a balanced chemical equation for the complete combustion of carbon and stated two physical and two chemical properties of the product formed as follows:
(i) C(s) + 02(g) → CO2(g)
(ii) I.  –   it is a gas at room temperature
          –   it is a colourless gas
          – it is odourless gas
          – it is denser than air
           –  it is soluble in water
             – it readily liquefies and solidifies/easily compressible

II.        – reacts with alkali to form trioxocarbonate(IV)
– reacts with (burning) Mg to form MgO
– reacts with red hot carbon to form CO
– it is weakly acidic in water
– it does not support combustion

In part (d), candidates named two crystalline and two amorphous carbon and one use of each of the named allotropes.
In part (e) (i), candidates were able to outline the process of manufacture of H2S04 by contract process. However some lost mark because of reversibility sign. The expected responses from candidates were:

 

In part (e)(ii), only few candidates correctly stated the function of H2S04 as dehydrating agent and precipitating/displacement agent in I and II respective.

ANSWER

n (a)(i), most of the candidates who attempted the question could not explain why water was referred to as universal solvent. They did not know that because it is polar, it can dissolve ionic solute as well as partially ionic/polar covalent substances.

In (a)(ii), candidates were able to give one chemical test for water thus:

Add water to anhydrous copper(II) tetraoxosulphate (VI), salt colour changes from white to blue
OR
Add water to anhydrous cobalt(II) chloride, colour changes from blue to pink.

In part (b)(i) and (ii), candidates could not write balanced equation for the oxidation half reaction, reduction half reaction and overall reaction nor calculate the volume of gas at the anode at s.t.p.

The expected responses from candidates were as stated below:

Ill. 2H20(l) → 2H2(g) + 02(g)

40H- (aq) + 4H+ (aq) →’2H2O(l) + 2H2(g) + 02(g)

(ii) Quantity of electricity = 1 x t
                                                  = 1.25 x 40 x 60
                                                 = 3000C
For 1 mole of O2 formed at the anode, 4e- are generated
Volume of O2 formed by 3000e = 1000 x 1 mole O2
                                                          96500    4
                                                    = 7.77 X 10-3 moles

1 mole of 02 occupies 22.4 dm3 at s.t.p
:. mole of 02 = 7.77 X 10-3 x 22.4 dm3

OR
From the equation at anode,
4 x 96500C = 22.4 dm3
3000C = 22.4 x 3000
4 x 96500
0.174dm3/174cm3
In part (c)(i), candidates could not balance the given chemical equation. The expected balanced chemical equation from candidates was

2Au(s) + 3CL2(g) →2AuCl3(s)

In part (c )(ii), only very few candidates could determine which of the reactants was in excess and the excess amount in I and II. The expected answers from candidates were as follows:

(ii) Amount of Au = 1.25
                                   197
                        0.00634 mole

Amount of CL2 1.744
                             71        = 0.0246 mole

From equation: 2 moles of Au = 3 moles of Cl2
0.00634 mole of Au = 0.00634 x 3
                                                2
                         = 0.00951 mole Cl2

Amount of Cl2 available = 0.0246 mole but amount required by all the
Au for the reaction = 0.00951
:. Cl2 is in excess

(iii) Excess amount of CL2 = amount available – amount used
                         = 0.0246 – 0.00951
                         = 0.015 mole

Alternative Method
(i) 2Au(s) + 3 Cl2(g) 2AuCl3(s)

(ii) 2 x 197g of Au = 3 x71g of Cl2
394g Au = 213g of Ci.,
1.2Sg = 213 x 1.25
                    394
           = 0.675g of Cl2
1.25g of Au will require 0.675g of Cl2
:. Cl2 is in excess

(iii) Excess amount of Cl2 = amount available – amount used
                    = 1.744 – 0.675
                          = 1.069g