B, neutralizes 25cm3 of 0.05M sulphuric acid solution?

Step 1: Write the balanced chemical equation for the neutralization reaction. Sulfuric acid ($\text{H}_2SO_4$) reacts with sodium hydroxide ($\text{NaOH}$) to produce sodium sulfate ($\text{Na}_2SO_4$) and water ($\text{H}_2\text{O}$). $H_2SO_4(aq) + 2NaOH(aq) \longrightarrow Na_2SO_4(aq) + 2H_2O(l)$ From the balanced equation, 1 mole of $\text{H}_2SO_4$ reacts with 2 moles of $\text{NaOH}$.

Step 2: Calculate the moles of sulfuric acid. Given: Volume of $\text{H}_2SO_4$ solution = 25 $\text{cm}^3$ = 0.025 L Concentration of $\text{H}_2SO_4$ solution = 0.05 M $Moles of H_2SO_4 = Molarity \times Volume (L)$ $Moles of H_2SO_4 = 0.05 mol/L \times 0.025 L$ $Moles of H_2SO_4 = 0.00125 mol$

Step 3: Calculate the moles of sodium hydroxide required for neutralization. Using the mole ratio from the balanced equation (1 mole $\text{H}_2SO_4$ : 2 moles $\text{NaOH}$): $Moles of NaOH = Moles of H_2SO_4 \times \frac{2 mol NaOH}{1 mol H_2SO_4}$ $Moles of NaOH = 0.00125 mol \times 2$ $Moles of NaOH = 0.0025 mol$

Step 4: Calculate the volume of sodium hydroxide solution required. Given: Concentration of $\text{NaOH}$ solution = 0.1 M $Volume of NaOH (L) = \frac{Moles of NaOH}{Molarity of NaOH}$ $Volume of NaOH (L) = \frac{0.0025 mol}{0.1 mol/L}$ $Volume of NaOH (L) = 0.025 L$

Step 5: Convert the volume to $\text{cm}^3$ (optional, but consistent with given units). $Volume of NaOH (cm^3) = 0.025 L \times \frac{1000 cm^3}{1 L}$ $Volume of NaOH (cm^3) = 25 cm^3$

The volume of 0.1 M sodium hydroxide solution required is $\boxed{0.025 L}$ or $\boxed{25 cm^3}$.