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 %
