What volume of a molar solution of tetraoxosulphate (VI) acid would exactly neutralize 25.0cm of potassium hydroxide solution containing 21.0g of the alkali in 250cm³? (K=39, H=1,0=16).

Tetraoxosulphate (VI) acid is $\text{H}_2\text{SO}_4$. Potassium hydroxide is $\text{KOH}$.

$$\text{H}_2\text{SO}_4(\text{aq}) + 2\text{KOH}(\text{aq}) \rightarrow \text{K}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$$

Given atomic masses: K = 39, O = 16, H = 1.

$$\text{Molar mass of KOH} = 39 + 16 + 1 = 56 \text{ g/mol}$$

Mass of $\text{KOH} = 21.0 \text{ g}$.

$$\text{Moles of KOH} = \frac{\text{Mass of KOH}}{\text{Molar mass of KOH}} = \frac{21.0 \text{ g}}{56 \text{ g/mol}} = 0.375 \text{ mol}$$

Volume of $\text{KOH}$ solution = 250 cm³ = 0.250 dm³.

$$\text{Concentration of KOH} = \frac{\text{Moles of KOH}}{\text{Volume of solution (dm}^3)} = \frac{0.375 \text{ mol}}{0.250 \text{ dm}^3} = 1.5 \text{ mol/dm}^3$$

Volume of $\text{KOH}$ used = 25.0 cm³ = 0.025 dm³.

$$\text{Moles of KOH used} = \text{Concentration of KOH} \times \text{Volume of KOH used}$$ $$\text{Moles of KOH used} = 1.5 \text{ mol/dm}^3 \times 0.025 \text{ dm}^3 = 0.0375 \text{ mol}$$

From the balanced equation, 1 mole of $\text{H}_2\text{SO}_4$ reacts with 2 moles of $\text{KOH}$.

$$\text{Moles of H}_2\text{SO}_4 = \frac{1}{2} \times \text{Moles of KOH used}$$ $$\text{Moles of H}_2\text{SO}_4 = \frac{1}{2} \times 0.0375 \text{ mol} = 0.01875 \text{ mol}$$

A "molar solution" means the concentration is 1 mol/dm³.

$$\text{Volume of H}_2\text{SO}_4 = \frac{\text{Moles of H}_2\text{SO}_4}{\text{Concentration of H}_2\text{SO}_4}$$ $$\text{Volume of H}_2\text{SO}_4 = \frac{0.01875 \text{ mol}}{1 \text{ mol/dm}^3} = 0.01875 \text{ dm}^3$$

Convert the volume to cm³:

$$\text{Volume of H}_2\text{SO}_4 = 0.01875 \text{ dm}^3 \times 1000 \text{ cm}^3/\text{dm}^3 = 18.75 \text{ cm}^3$$

The volume of a molar solution of tetraoxosulphate (VI) acid required is $\boxed{\text{18.75 cm}^3}$.