Calculate the number of hydrogen ions in 1.0 dm³ of 0.02 moldm⁻³ sulfuric acid.

Tetraoxosulphate (VI) acid is sulfuric acid, $\text{H}_2\text{SO}_4$. It is a strong acid and dissociates completely in water to produce hydrogen ions.
The dissociation equation is:

$$\text{H}_2\text{SO}_4 \rightarrow 2\text{H}^+ + \text{SO}_4^{2-}$$

This shows that 1 mole of $\text{H}_2\text{SO}_4$ produces 2 moles of $\text{H}^+$ ions.

Given concentration ($C$) = $0.02 \text{ moldm}^{-3}$
Given volume ($V$) = $1.0 \text{ dm}^3$
The number of moles ($n$) is calculated as:

$$n(\text{H}_2\text{SO}_4) = C \times V$$ $$n(\text{H}_2\text{SO}_4) = 0.02 \text{ moldm}^{-3} \times 1.0 \text{ dm}^3 = 0.02 \text{ mol}$$

From the dissociation equation, 1 mole of $\text{H}_2\text{SO}_4$ produces 2 moles of $\text{H}^+$ ions.

$$n(\text{H}^+) = 2 \times n(\text{H}_2\text{SO}_4)$$ $$n(\text{H}^+) = 2 \times 0.02 \text{ mol} = 0.04 \text{ mol}$$

Given Avogadro's number ($N_A$) = $6.02 \times 10^{23} \text{ mol}^{-1}$
The number of ions is calculated as:

$$\text{Number of H}^+ \text{ ions} = n(\text{H}^+) \times N_A$$ $$\text{Number of H}^+ \text{ ions} = 0.04 \text{ mol} \times 6.02 \times 10^{23} \text{ mol}^{-1}$$ $$\text{Number of H}^+ \text{ ions} = (4 \times 10^{-2}) \times (6.02 \times 10^{23})$$ $$\text{Number of H}^+ \text{ ions} = 24.08 \times 10^{21}$$ $$\text{Number of H}^+ \text{ ions} = 2.408 \times 10^{22}$$

Rounding to two significant figures, this is $2.4 \times 10^{22}$.

Comparing this to the given options:

(a) $1.2 \times 10^{22}$

(b) $1.2 \times 10^{23}$

(c) $2.4 \times 10^{22}$

(d) $2.4 \times 10^{23}$

The calculated value matches option (c).

The correct option is c).
$\boxed{\text{c) } 2.4 \times 10^{22}}$