(a)(i) Name two gases that could be used to perform the fountain experiment.
(ii) State the physical property which makes the gases suitable for the experiment in (a)(i)
(b)(i) Define each of the following terms: I. solubility; II. saturated solution.
(ii) State two factors that affect the solubility of a solid in a liquid.
(iii) A salt Z of mass 10.2 g was dissolved in 15.4 cm\(^3\) of distilled water at 40°C. Calculate the solubility of Z in moldm\(^3\) at 40°C. [Mr (Z) = 331].
(c)(i) Town water supplies that have passed through iron pipes contain P and Q ions. In the presence of air, P ions are slowly converted to Q ions.
I. Identify P and Q ions. II. Write a balanced equation for the reaction between P ions, hydrogen ions and oxygen to give Q ions and water.
(ii) Explain briefly a test to confirm the purity of water.
(iii) State the effect of:
I. boiling a temporally hard water. II. adding sodium trioxocarbonate (IV) crystals to permanent hard water;
(iv) Write a balanced equation for the process in (c)(iii)I.
(a)(i) Gases that could be used to perform the fountain experiment: –Hydrogen chloride. Ammonia
(ii) Physical property which makes the gases suitable for the experiment: –The gases are highly/very soluble in water.
(b)(i) Definition of Solubility and Saturated Solution: I. Solubility is the maximum amount of soluble in mole or grams that will dissolve in 1 dm\(^3\) of the solvent at that temperature or solubility is the amount of solute that will saturate 1 dm\(^3\) of the solvent at that temperature.
II. Saturated solution is a solution that contains as much solute that can dissolve at a particular temperature (in the presence of undissolved solute particles).
(ii) Factors that affect the solubility of a solid in a liquid: –Temperature. –Nature of the solute. --Nature of the solvent.
(iii) A salt Z of mass 10.2 g was dissolved in 15.4 cm\(^3\)of distilled water at 40°C. Calculate the solubility of Z in moldm\(^3\) at 40°C. Calculate the solubility of Z in moldm\(^3\) at 40°C
1000cm of water = \(\frac{10.2 \times 1000}{15.4}\)
= 662.34g
Solubility in moldm\(^3\) = \(\frac{662.34}{331}\)
= 2.0moldm\(^3\)
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