Assumption: We will assume the average volume of solution A (H2SO4) used in the titration (V_A) is 20.00 cm3.

To solve this problem, we first need to address the missing titration data. Since no experimental readings are provided for the titration, we will assume a typical average volume of solution A used.

Assumption: We will assume the average volume of solution A (H$_2$SO$_4$) used in the titration ($V_A$) is \textbf{20.00 cm^3}.

i) Concentration of B in mol/dm$^3$

Step 1: Calculate the molar mass of Sodium hydroxide (NaOH).
The relative atomic masses are H=1.00, O=16.0, Na=23.0.

$$\text{Molar mass of NaOH} = 23.0 + 16.0 + 1.00 = 40.0 \text{ g/mol}$$

Step 2: Calculate the number of moles of NaOH in 1 g.

$$\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1.00 \text{ g}}{40.0 \text{ g/mol}} = 0.0250 \text{ mol}$$

Step 3: Calculate the concentration of solution B (NaOH) in mol/dm$^3$.
The volume of solution B is 250 cm$^3$, which is 0.250 dm$^3$.

$$\text{Concentration of B} (C_B) = \frac{\text{Moles of NaOH}}{\text{Volume of solution B}} = \frac{0.0250 \text{ mol}}{0.250 \text{ dm}^3} = \boxed{\text{0.100 mol/dm}^3}$$

ii) Concentration of A in mol/dm$^3$

Step 4: Write the balanced chemical equation and determine the mole ratio.
The balanced equation is:

$$\text{H}_2\text{SO}_4 \text{(aq)} + 2\text{NaOH (aq)} \longrightarrow \text{Na}_2\text{SO}_4 \text{(aq)} + 2\text{H}_2\text{O (l)}$$

From the equation, 1 mole of H$_2$SO$_4$ reacts with 2 moles of NaOH. So, $n_A = 1$ and $n_B = 2$.

Step 5: Use the titration formula to calculate the concentration of A (H$_2$SO$_4$).
Given:
$C_B = 0.100 \text{ mol/dm}^3$ (from part i)
$V_B = 25 \text{ cm}^3 = 0.025 \text{ dm}^3$
$V_A = 20.00 \text{ cm}^3 = 0.02000 \text{ dm}^3$ (assumed average volume)
The titration formula is:

$$\frac{C_A V_A}{n_A} = \frac{C_B V_B}{n_B}$$

Rearranging to solve for $C_A$:

$$C_A = \frac{C_B V_B n_A}{V_A n_B} = \frac{(0.100 \text{ mol/dm}^3) \times (0.025 \text{ dm}^3) \times 1}{(0.02000 \text{ dm}^3) \times 2}$$ $$C_A = \frac{0.0025}{0.04000} = \boxed{\text{0.0625 mol/dm}^3}$$

iii) Number of hydrogen ions in 1.0 dm$^3$ of A

Step 6: Calculate the moles of H$_2$SO$_4$ in 1.0 dm$^3$ of solution A.
The concentration of A ($C_A$) is 0.0625 mol/dm$^3$.

$$\text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 0.0625 \text{ mol/dm}^3 \times 1.0 \text{ dm}^3 = 0.0625 \text{ mol}$$

Step 7: Determine the moles of hydrogen ions (H$^+$) produced.
Sulfuric acid (H$_2$SO$_4$) is a strong acid and dissociates completely in water:

$$\text{H}_2\text{SO}_4 \text{(aq)} \longrightarrow 2\text{H}^+ \text{(aq)} + \text{SO}_4^{2-} \text{(aq)}$$

Each mole of H$_2$SO$_4$ produces 2 moles of H$^+$ ions.

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