Given average titre (volume of acid A used) = 12.30 cm3

Solution B contains $1\text{ g}$ of $\text{NaOH}$ in $250\text{ cm}^3$.

$$\text{Moles of NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{1\text{ g}}{40.0\text{ g/mol}} = 0.025\text{ mol}$$ $$\text{Volume of solution B} = 250\text{ cm}^3 = 0.250\text{ dm}^3$$ $$\text{Concentration of B} = \frac{\text{moles}}{\text{volume}} = \frac{0.025\text{ mol}}{0.250\text{ dm}^3} = \boxed{0.10\text{ mol dm}^{-3}}$$

Part (b)(ii): Calculate the concentration of A in $\text{mol dm}^{-3}$.

The balanced chemical equation is: $\text{H}_2\text{Y} + 2\text{NaOH} \longrightarrow \text{Na}_2\text{Y} + 2\text{H}_2\text{O}$
From the equation, the mole ratio of $\text{H}_2\text{Y}$ to $\text{NaOH}$ is $1:2$.
Given:
$V_A = 12.30\text{ cm}^3 = 0.01230\text{ dm}^3$
$V_B = 25\text{ cm}^3 = 0.025\text{ dm}^3$
$C_B = 0.10\text{ mol dm}^{-3}$ (from Step 2)
Using the titration formula $\frac{C_A V_A}{n_A} = \frac{C_B V_B}{n_B}$:

$$C_A = \frac{n_A \times C_B \times V_B}{n_B \times V_A} = \frac{1 \times 0.10\text{ mol dm}^{-3} \times 0.025\text{ dm}^3}{2 \times 0.01230\text{ dm}^3}$$ $$C_A = \frac{0.0025}{0.0246}\text{ mol dm}^{-3} \approx \boxed{0.1016\text{ mol dm}^{-3}}$$

Part (b)(iii): Calculate the number of hydrogen ions in $1\text{dm}^3$ of A.

Concentration of A ($C_A$) = $0.1016\text{ mol dm}^{-3}$.
In $1\text{ dm}^3$ of solution A, moles of $\text{H}_2\text{Y} = 0.1016\text{ mol}$.
Since $\text{H}_2\text{Y}$ is a diprotic acid, each molecule produces 2 $\text{H}^+$ ions.

$$\text{Moles of H}^+\text{ ions in } 1\text{ dm}^3 = 2 \times 0.1016\text{ mol} = 0.2032\text{ mol}$$

Using Avogadro's number ($6.02 \times 10^{23}\text{ ions/mol}$):

$$\text{Number of H}^+\text{ ions} = 0.2032\text{ mol} \times 6.02 \times 10^{23}\text{ ions/mol} \approx \boxed{1.22 \times 10^{23}\text{ ions}}$$

Part (b)(iv): Calculate the volume of distilled water that should be added to $25\text{cm}^3$ of solution B in order to dilute it twice.