Determine the formula of the hydrated salt and calculate the concentration of the resulting salt solution.

Assume a 100 g sample of the hydrated salt.
Mass of Iron ($\text{Fe}$) = 20.2 g
Mass of Sulphur ($\text{S}$) = 11.5 g
Mass of Oxygen ($\text{O}$) = 23.0 g
Mass of Water ($\text{H}_2\text{O}$) = 45.3 g

Moles of $\text{Fe} = \frac{20.2 \text{ g}}{56 \text{ g/mol}} = 0.3607 \text{ mol}$
Moles of $\text{S} = \frac{11.5 \text{ g}}{32 \text{ g/mol}} = 0.3594 \text{ mol}$
Moles of $\text{O} = \frac{23.0 \text{ g}}{16 \text{ g/mol}} = 1.4375 \text{ mol}$
Moles of $\text{H}_2\text{O} = \frac{45.3 \text{ g}}{18 \text{ g/mol}} = 2.5167 \text{ mol}$

For $\text{Fe}: \frac{0.3607}{0.3594} \approx 1$
For $\text{S}: \frac{0.3594}{0.3594} = 1$
For $\text{O}: \frac{1.4375}{0.3594} \approx 4$
For $\text{H}_2\text{O}: \frac{2.5167}{0.3594} \approx 7$

The ratio of $\text{Fe} : \text{S} : \text{O} : \text{H}_2\text{O}$ is $1:1:4:7$.
This indicates the anhydrous part is $\text{FeSO}_4$ and there are 7 molecules of water of crystallization.

The formula of the hydrated salt is $\boxed{\text{FeSO}_4 \cdot 7\text{H}_2\text{O}}$.

13. ii) Calculate the concentration of the resulting salt solution in moles per litre.

Given mass of hydrated salt = 6.95 g
Given molecular mass of the salt = 278 g/mol

$$\text{Moles of salt} = \frac{\text{Mass}}{\text{Molecular mass}} = \frac{6.95 \text{ g}}{278 \text{ g/mol}} = 0.025 \text{ mol}$$

Given total volume = 250 $\text{cm}^3 = 0.250 \text{ dm}^3$

$$\text{Concentration} = \frac{\text{Moles of salt}}{\text{Volume of solution in dm}^3} = \frac{0.025 \text{ mol}}{0.250 \text{ dm}^3} = 0.1 \text{ mol/dm}^3$$

The concentration of the resulting salt solution is $\boxed{\text{0.1 M}}$.

14. (i) Calculate the mass in grams of lead (II) iodide formed.

Given volume of iodide ions = 300 $\text{cm}^3 = 0.300 \text{ dm}^3$
Given molarity of iodide ions = 0.1 M

$$\text{Moles of } \text{I}^- = \text{Molarity} \times \text{Volume} = 0.1 \text{ mol/dm}^3 \times 0.300 \text{ dm}^3 = 0.030 \text{ mol}$$

The balanced equation is: $\text{Pb}^{2+} (aq) + 2\text{I}^- (aq) \rightarrow \text{PbI}_2 (s)$
From the equation, 2 moles of $\text{I}^-$ react to produce 1 mole of $\text{PbI}_2$.

$$\text{Moles of } \text{PbI}_2 = \frac{1}{2} \times \text{Moles of } \text{I}^- = \frac{1}{2} \times 0.030 \text{ mol} = 0.015 \text{ mol}$$

Atomic mass of $\text{Pb}$ = 207
Atomic mass of $\text{I}$ = 127

$$\text{Molar mass of } \text{PbI}_2 = 207 + (2 \times 127) = 207 + 254 = 461 \text{ g/mol}$$

$$\text{Mass of } \text{PbI}_2 = \text{Moles} \times \text{Molar mass} = 0.015 \text{ mol} \times 461 \text{ g/mol} = 6.915 \text{ g}$$

The mass of lead (II) iodide formed is $\boxed{\text{6.915 g}}$.

14. (ii) Identify the colour of the product formed in (i).
The product formed is lead (II) iodide ($\text{PbI}_2$).
The colour of lead (II) iodide is $\boxed{\text{yellow}}$.