(a (i) Explain why water is referred to as a universal solvent.
(ii) Give one chemical test for water. H\(_2\)SO\(_4\)
(b) A current of 1.25A was passed through an electrolytic cell containing dil. for 40 minutes.
(i) Write (ii) Calculate the volume of gas produced at the anode at s.t.p. (1F = 96,500 C, Molar Volume of gas at s.t.p. = 22.4 dm\(^3\) mol\(^{-3}\) ]
(C) Consider the reaction represented by the following equation: Au\(_{(s)}\) + Cl\(_{2(g)}\) \(\to\) AuCl\(_{3(s)}\)
(I) Balance the equation
(ii) If 1.2509 of Au and 1.744g of Cl\(_2\) were mixed
I. determine which of the reactants is in excess II. calculate the excess amount [Au = 197.0, CI = 35.5]
Explanation
(a)(i) Water is a universal solvent because it can dissolve many substances. water is polar and as a result, it can dissolve ionic solutes as well as partially ionic/polar covalent substances
(ii) Add water to anhydrous copper
(II) tetraoxosulphate (VI) salt, colour changes from white to blue. OR Add water to anhydrous cobalt
(II) ch!oric!e, colour changes from blue to pink.
(b)(i) I. OH\(^-{_(aq)}\) \(\to\) 2H\(_2\)O\(^-{_(l)}\) + O\(_{2(g)}\) + 4e\(^-\)
II. 2H\(^+_{(aq)}\) + 2e\(^-\) --> H\(_{2(g)}\)
OR 4H\(^+_{(aq)}\) + 4e\(^-\) --> 2H\(_{2(g)}\)
Ill. 2H\(_2\)O\(_{(l)}\) -> OH\(_{2(g)}\) + O\(_{2(g)}\)
OR 4OH\(^-_{(aq)}\) + 4H\(^+_{(g)}\) --> 2H\(_2\)O\(_{(l)}\) + 2H\(_{2(g)}\) + O\(_{2(g)}\)
(ii) Quantity of electricity = 1 x t = 1.25 x 40 x 60 = 3000C
For 1 mole of O\(_2\) formed at the anode, 4e\(^-\) are generated Volume of O\(_2\) formed by 3000C = \(\frac{3000}{96500}\) x \(\frac{1}{4}\) mole O\(_2\)
= 7.77 x 10\(^{-3}\) moles
1 mole O\(_2\) occupies 22.4 dm\(^{3}\) at s.t.p. mole of O\(_2\) = 7.77 x 10\(^{-3}\) x 22.4 dm\(^{3}\) = 0.174dm\(^{3}\)/174 cm\(^{3}\)
OR From the equation at anode; 4 x 96500C = 22.4 dm\(^{3}\)
3000C = \(\frac{22.4 \times 3000}{4 \times 96500}\)
= 0.174dm\(^{3}\) / 174 cm\(^{3}\)
(c)(i) 2Au\(_{(s)}\) + 3Cl\(_{2(g)}\) \(\to\) 2AuCI\(_{3(s)}\)
(ii) Amount of Au= \(\frac{1.2}{197}\) = 0.00634 mole
Amount of Cl\(_2\) = \(\frac{1.744}{71}\) = 0.0246 mole.
From equation: 2 moles of Au = 3 moles of Cl\(_{(2)}\)
.-. 0.00634 mole of Au = \(\frac{0.00634 \times 3}{2}\) = 0.00951 mole
Amount of CI\(_2\) available = 0.0246 mole but amount required by all the Au for the reaction = 0.00951
CI\(_2\) is in excess.
(iii) Excess amount of CI\(_2\) = amount available - amount used = 0.0246 - 0.00951 = 0.015 mole.