This question was attempted by very few candidates and the performance was fair.
In(a)(i), candidates correctly stated the procedures that would be carried out in the laboratory for each of I – III as follows:
I. Add washing soda (Na2CO3 10H2O)/pass it through an ion-exchange
resin add calculated amount of slaked lime (Ca(OH)2).
III. Filtration (accept decantation)/centrifuging.
In (a) (ii), candidates stated correctly that alum is for coagulation of fine/colloidal particles and chlorine is for killing bacteria in a water treatment plant.
In(a)(iii), candidates correctly gave two advantages of soft water over hard water from among the following:
- it does not waste soap/lathers readily with soap
- it does not cause furring/scales/deposits in kettles and boilers
- it is cheaper when used in dyeing and tanning industries.
In (b)(i), candidates gave the name of a metallic oxide which decreases in mass without dissolving in it when exposed to the atmosphere as calcium oxide/copper(II) oxide/magnesium oxide.
In (b)(ii), most of the candidates knew that the phenomenon is hygroscopy.
In (b)(iii), candidates could not write a balanced chemical equation for the reaction between the metallic oxide and dilute trioxonitrate (V) acid. The expected answer was
CaO(s) + 2HNO3(aq), → Ca(NO3)2(aq) + H2O
CuO(s) + 2HNO3(aq), → Cu(NO3)2(aq) + H2O
In (c)(i), candidates could not identify W, X, Y and Z.
The expected answers were:
W - oxygen/ O2(g)
X - flat bottom flask
Y - conc H2SO(4)/fused CaCl2/U-tube
Z - mercury
In (c)(ii), candidates knew that manganese (IV) oxide is acting as a catalyst.
In (c)(iii), candidates could not give a balanced chemical equation for the preparation of the gas W
2H2O2(aq) 2H2O(l) + O2(g)
In (c)(iv), candidates knew that W cannot be collected by the displacement of air because it has nearly the same density with air.
In (d), most candidates did not know that nitrogen is diatomic hence lost marks for using N = 14 instead of N2 = 28. The expected solution to the question is as follows:
Number of moles of nitrogen = mass = 2.8
molar mass 28
= 0.10 moles
1 mole of nitrogen = 6. 02 x 1023 moles
0.1 moles of nitrogen will contain 6.02 x1023 x 0.10
= 6.02 x 1022 particles