Describe the electrolysis of concentrated sodium chloride solution using graphite electrodes, including the role of graphite, products, and calculations related to hydrogen gas.

Here's the solution to your question:

a) Graphite is used as an electrode because it is a good conductor of electricity and it is inert (unreactive) with the electrolyte and products of the reaction.

b)

(i) The electrolyte is concentrated sodium chloride solution. The ions present include:
$\text{Na}^+$ and $\text{Cl}^-$

(ii) The solution would be alkaline.
Explanation: During the electrolysis of concentrated sodium chloride, chloride ions ($\text{Cl}^-$) are discharged at the anode, and hydrogen ions ($\text{H}^+$) from water are discharged at the cathode. This leaves an excess of hydroxide ions ($\text{OH}^-$) and sodium ions ($\text{Na}^+$) in the solution, forming sodium hydroxide ($\text{NaOH}$), which is a strong alkali.
The reaction at the cathode is: $2\text{H}_2\text{O}(l) + 2\text{e}^- \to \text{H}_2(g) + 2\text{OH}^-(aq)$

c)

(i) To calculate the number of moles in 0.025 g of hydrogen gas ($\text{H}_2$):
Molar mass of $\text{H}_2 = 2 \times 1.0 \text{ g/mol} = 2.0 \text{ g/mol}$

$$\text{Moles of gas} = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Moles of gas} = \frac{0.025 \text{ g}}{2.0 \text{ g/mol}}$$ $$\text{Moles of gas} = \boxed{\text{0.0125 mol}}$$

(ii) To calculate the volume of hydrogen gas produced at room temperature and pressure (RTP), where 1 mole of gas = 24 dm$^3$:

$$\text{Volume of gas} = \text{Moles} \times \text{Molar volume at RTP}$$ $$\text{Volume of gas} = 0.0125 \text{ mol} \times 24 \text{ dm}^3/\text{mol}$$ $$\text{Volume of gas} = \boxed{\text{0.3 dm}^3}$$