(a) Define the following terms:(i) Saturated solution; (ii) Solubility.
(b) In an experiment to determine the solubility of a given salt Y, the following data were provided:
Mass of dry empty dish = 7.16 g
Mass of dish + saturated solution of salt Y = 17.85 g
Mass of dish + salt Y = 9.30 g Temperature of solution = °C
Molar mass of salt Y = 100
Density of solution Y = 1.00 g cm\(^{-3}\) Calculate the solubility of salt Y in
(i) g dm\(^{-3}\) of solution, (ii) mol dm\(^{-3}\) of solution.
(c) State the type of bond broken on melting each of the following substances: (i) NaCI\(_{(s)}\) (ii) CO\(_{2(s)}\) (iii) SiO\(_{2(s)}\) (iv) Al\(_{(s)}\)
(d) Explain the following observations: (i) the chemical reactivity of alkali metals increases down the group; (ii) Mg has higher melting point than Na; (iii) K is a better reducing agent than Na.
(e)(i) What are isotopes? (ii) Lithium exists as \(^6_3\)Li and \(^7_3\)Li in the ratio 2:25. Calculate the relative atomic mass of the lithium.
Explanation
(a)(i) A saturated solution is one that will dissolve no more of the solute at a particular temperatuie in the presence of excess solute.
(ii) Solubility is the,maximum amount of solute (number of moles/mass) that dissolves in 1dm\(^3\) of solvent at a particular temperature.
(b) Mass of saturated solution = 17.85 — 7.16 = 10.69
Mass of dry salt = 9.30 — 7.16 = 2.14
Volume of solution = 10.69cm\(^3\)
Since volume of solution = \(\frac{mass}{density}\)
and density = 1.0g/dm\(^3\)
(i) Solubility in g/dm\(^3\)
10.69cm\(^3\) of solution Y contains 2.14g of salt Y
1cm\(^3\) of solution Y contains \(\frac{2.14 }{10.69}\)
1000cm\(^3\) = \(\frac{2.14 \times 1000}{10.69}\) = 200.19g/dm\(^3\)
(ii) Solubility in mol/dm\(^3\) = \(\frac{\text{solubility in g/dm}^3}{\text{molar mass of salt Y}}\)
= \(\frac{200.19}{100}\)
= 2.00 mol/dm\(^3\)
(c) NaCI — Ionic bond of electrovalent bond
CO\(_{2}\) — van der Waals forces
SiO — Covalent bond Al\(_{(s)}\) — Metallic bond.
(d)(i) Reactivity is due to loss of electrons, i.e the ease of electron loss, hence ionization energy decreases down the group.
(ii) Mg has more valence electrons contributing to the stronger binding energy compared to Na.
(iii) Potassium loses electron more readily than sodium, has lower ionization energy than sodium.
(e)(i) Isotopes are atoms of the same element with the same atomic number but different mass number.
(ii) Li = \(\frac{2 \times 6}{27}\) + \(\frac{25 \times 7}{27}\)
= 6.925