Question

2. (a)(ii) Charles' law states that the volume of a given mass of a gas is directly proportional to the absolute temperature, provided that the pressure remains constant.
(iii) V∝T, V=KT, K=V/T, so V1/T1 = V2/T2.
(iv) A given mass of a gas occupied 150cm³ at 27°C and a pressure of 1.013×10⁵Nm⁻². Calculate the temperature at which its volume will be doubled at the same pressure.
(v) Arrange the three states of matter in order of increasing: I. Kinetic energy; II. forces of cohesion.
(b)(i) State Le Chatelier's principle. (ii) A metal M forms two oxides containing 11.1% and 20% of oxygen. Show that these figures agree with the law of multiple proportion.
(c) The table below shows the physical properties of substances A, B and C.
Substance | Melting point/°C | Boiling point/°C | Solubility in water at 25°C
A | 30 | 117 | Insoluble
B | 31 | 160 | Insoluble
C | 861 | 1200 | Soluble
(i) If A and B are miscible when melted and B and C react when heated, describe how a mixture of A, B and C could be separated.
(ii) When 25.25g of the mixture A, B and C was separated, 7.52g of A and 8.48g of B were recovered. Assuming there was no loss of components during the separation, calculate the percentage by mass of C in the mixture.

Accepted Answer

(iv) P1=P2=1.013×10⁵Nm⁻² (constant pressure), so Charles' law applies. T1=300K, V1=150cm³, V2=300cm³ (doubled). T2 = (V2/V1)×T1 = (300/150)×300 = 600K = 327°C.
(v) Increasing kinetic energy: Solid<Liquid<Gas. Increasing forces of cohesion: Gas<Liquid<Solid.

(b)(i) Le Chatelier's principle states that when a physical constraint is imposed on a chemical system at equilibrium, the equilibrium position adjusts so as to neutralize the effect.
(ii) Let 100% = 100g. For the first oxide: mass of oxygen = 11.1g, so mass of metal = 100−11.1 = 88.9g. Fixing a mass of oxygen (say 50g): (88.9/11.1)×50 = 400g of the metal. For the second oxide: mass of oxygen = 20g, mass of metal = 100−20 = 80g. (80/20)×50 = 200g of the metal. Ratio of the metal in the two oxides = 400:200 = 2:1, a simple whole-number ratio, confirming the law of multiple proportion.

(c)(i) Water should be added to the mixture. Since C is soluble, it dissolves. B and A are then filtered off as residue, since they are not soluble. The dry form of C can be obtained by evaporating the filtrate. The residue, containing A and B, should be heated until it melts. A and B are then separated using simple distillation, since their boiling points are far apart (A boils at 117°C while B boils at 160°C).
(ii) Mass of mixture = (mass of A)+(mass of B)+(mass of C). 25.25g = 7.52g+8.48g+(mass of C). Mass of C = 25.25−16 = 9.25g. % of C in the mixture = (9.25/25.25)×100% = 36.63%.

WAEC Chemistry 2012 Theory — Question 2

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