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Chemistry Paper 2,Nov/Dec 2010  
Questions:   1 2 3 4 5 6 7 8   Main
General Comments
Weakness/Remedies
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Question 8

(a)     When 28.0cm3 of oxygen gas at 27°C and a constant pressure was immersed in an ice bath at OOC, the volume of oxygen decreased.

(i) Which gas law is illustrated by this observation?
(ii) State the gas law.
(iii) Sketch a graph to illustrate this law.
(iv) Calculate the new volume of oxygen. [9 marks]

(b) Explain why:
(i) it is necessary to rinse a garment thoroughly with water after bleaching with chlorine;
(ii) dilute HNO3 cannot be used to prepare hydrogen gas;
(iii) CO2 is employed in fire extinguisher;



(iv)     C02 prepared from dilute HCl and CaC03, is passed through a solution of KHC03, before
the gas is dried.
(c) (i) Name the two products formed when chlorine-water is exposed to sunlight.

(ii) Give the main product of the reaction of chlorine with cold dilute NaOH.

(iii)State the variation of each of the following properties down the halogen group:
I.           boiling point;
II.           oxidizing ability.

Give reason for each of your answers [8 marks]

                                                

_____________________________________________________________________________________________________
OBSERVATION

The question was popular with the candidates and the performance was poor.

In (a)(i), candidates knew that the law illustrated was Charles' law.

n (a)(ii), they correctly stated that the volume of a given mass of gas is directly proportional to its temperature in (Kelvin) provided that the pressure remains constant was the statement of the law.

In(a)(iii), only few candidates correctly sketched a graph to illustrate the law as follows:
         



In(a)(iv), most candidates applied the law to calculate the new volume of oxygen thus:


V1 = 28.0cm3
T1 = 270C = 27 x 273 = 300k
T2 = 00C + 273 = 273k
V2 = ?
V1/T1 = V2/T2
V2 = V1T2/T2 = 28 x 273/300
= 25.48cm3


In part (b), candidates could not give the required explanation in (i)-iv). The expected answer for each of (i) - Iv) was as follows:
      (i)        Chlorine bleaches item, (example garment) as dilute acid
                  During rinsing the water removes acid that can damage the material
      (ii)        Dilute HN03 oxidizes the hydrogen formed immediately to water
      (iii)        CO2 is heavier than air and does not support combustion
      (iv)       To remove the traces of acid fumes/HCl.

In (c)(i), candidates named oxygen and hydrochloric acid as the two products formed when chlorine water is exposed to sunlight. However, some of them lost the marks because they wrote O2 and
HCl which were not names but formula.

In (c)(ii), only very few candidates knew that NaOCI/NaCI was the main product of chlorine with cold dilute NaOH.
In(c)(iii), candidates could not state the variation of boiling point and oxidizing ability among the
halogen nor give reason for the variation.

The expected answer from candidates were as follow:
           I.          Boiling point increases down the halogen group due to increase in molecular mass/van der Waals forces

           II.          Oxidizing ability of the halogen decreases down the group due to decrease in effective nuclear charge as a result of increase in atomic size/decrease in electronegativity.




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