|
Question 1
|
(a) Define the following terms:
(i) Saturated solution;
(ii) Solubility. 100
(b)In an experiment to determine the solubility of a given salt Y, the following data were provided:
Mass of dry empty dish = 7.16g = 17.85g
Mass of dish + salt Y = 9.30g
Temperature of solution = 20oC
Molar mass of salt Y = 100
Density of solution Y = 1.00gcm-3
Calculate the solubility of salt Y in
(i) g dm-30f solution,
(ii) mol dm-3 of solution.
[6 marks]
(c) State the type of bond broken on melting each of the following substances:
NaCl(s);
CO2(s);
SiO2(s);
Al(s) [4 marks]
(d) Explain the following observations;
(i) the chemical reactivity of alkali metals increases down the group;
(ii) Mg has a higher melting point than Na;
(iii) K is a better reducing agent than Na. [6 marks]
(e) (i) What is isotopes?
(ii) Lithium exists as 6 Li and 7 Li in the ratio 2:25. Calculate the
3 3
Relative atomic mass of the lithium. [5 marks]
|
| _____________________________________________________________________________________________________ |
|
This question was attempted by majority of the candidates ,and the performance was poor.
In (a), most candidates correctly defined saturated solution and solubility as a solution that will not dissolve any more of the solute at a particular temperature in the presence of excess solute and the maximum amount of solute (mol/g) that dissolves in Idm3 of solvent at a particular temperature in (i) and (ii) respectively. However, some of them
204 lost marks because they did not know the importance of the temperature as a parameter in the definition.
In (b), only few candidates were able to correctly calculate the solubility of salt in gdm- 3 and
moldm-3 of solution in (i) and (ii) respectively thus:
mass of saturated solution = (17.85 -7.16) = 10.69g
mass of dry salt = (9.30-7.16) = 2,14g
Volume of solution = mass of solution since density = 1.0gcm-3
= 10.69cm3
(i) Solubility in g/dm3
10.69cm3 of solution Y contains 2.14g of salt Y
lcm3 of solution Y contains 2.14 gcm-3
10.69
1000cm3 = 2.14 x 1000
10.69
= .200.19gdm-3 (Accept 200.197 dgm-3)
(ii) Solubility in moldm·3 =
solubility in gdm-3
Molar mass of salt Y
= 200.19
100
= 2.22moldm-3
In (c)(i)-(iv), candidates correctly stated the type of bond broken on melting each of the substances as ionic, van der Waal forces, covalent and metallic respectively. Some of them however confused the type of bond broken in CO2(g) as covalent. They did not know that
CO 2(s) is a molecular solid.
In (d), a good number of the candidates could not give exp1ariation to the observations in
(i) - (iii). They did not know that
(i) Reactivity is due to loss of electrons/ease of electron loss, hence ionization energy decreases down the group.
(ii) Mg has more valence electron contributing to the stronger binding . energy compared to Na.
(iii) Potassium loses electron more. readily than sodium/K has lower ionization energy than Na. Ease of loss of electrons by an atom is a measure of its reducing ability.
In (e)(i), candidates correctly gave· the meaning of isotopes as atoms of the same element with the same atomic number but different mass number. However, a few of them confused "atoms" for "elements" and hence lost the marks
In (ii), majority of the candidates were able to calculate the relative atomic mass of lithium thus:
Li = 2x6. + 25x7
27 27
= 6.925 (Accept 6.93) |
|
|
|
