Explain the decolorization of chlorine water in sunlight and the preparation of lead sulfate.

*a) The diagram shows sunlight acting on chlorine water, which causes the decomposition of hypochlorous acid ($\text{HClO}$) present in chlorine water to produce oxygen gas.

The colourless gas X is oxygen.

*b) Chlorine water is yellow due to dissolved chlorine. In the presence of sunlight, the chlorine reacts with water to form hypochlorous acid ($\text{HClO}$) and hydrochloric acid ($\text{HCl}$). The hypochlorous acid then decomposes to produce oxygen gas and hydrochloric acid. The removal of the dissolved chlorine and its conversion into other compounds causes the solution to decolorize.

To prepare lead sulphate ($\text{PbSO}_4$) from solid potassium sulphate ($\text{K}_2\text{SO}_4$), solid lead carbonate ($\text{PbCO}_3$), dilute nitric acid ($\text{HNO}_3$), and distilled water:

1. React solid lead carbonate with dilute nitric acid to form soluble lead nitrate solution: $$\text{PbCO}_3\text{(s)} + \text{2HNO}_3\text{(aq)} \rightarrow \text{Pb(NO}_3)_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$
2. Filter the solution to remove any unreacted solid lead carbonate.
3. Add potassium sulphate solution to the lead nitrate solution. Lead sulphate, an insoluble salt, will precipitate out: $$\text{Pb(NO}_3)_2\text{(aq)} + \text{K}_2\text{SO}_4\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)} + \text{2KNO}_3\text{(aq)}$$
4. Filter the mixture to collect the solid lead sulphate precipitate.
5. Wash the precipitate with distilled water to remove soluble impurities (potassium nitrate).
6. Dry the lead sulphate precipitate.

*a) Boyle's Law states that for a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure. Mathematically, $P_1V_1 = P_2V_2$.

*b) Given:
$V_1 = 250 \text{ cm}^3$
$T_1 = 27^\circ\text{C}$
$P_1 = 750 \text{ mmHg}$
$T_2 = 41^\circ\text{C}$
$P_2 = 750 \text{ mmHg}$

Convert temperatures to Kelvin:
$T_1 = 27^\circ\text{C} + 273 = 300 \text{ K}$
$T_2 = 41^\circ\text{C} + 273 = 314 \text{ K}$

Since the pressure is constant ($P_1 = P_2$), we can use Charles's Law:

Substitute the given values:

Solve for $V_2$:

Rounding to three significant figures:

*a) • A deliquescent substance is a substance that absorbs moisture from the atmosphere and dissolves in it to form a solution.

• An efflorescent substance is a crystalline substance that loses its water of crystallization to the atmosphere when exposed to air, turning into a powder.

*b) One use of hygroscopic substances in the laboratory is as drying agents (desiccants) to remove moisture from gases or liquids.

*a) Allotropy is the property of some chemical elements to exist in two or more different forms, known as allotropes, in the same physical state. These allotropes have different physical properties and often different chemical properties due to differences in their atomic arrangement or bonding.