How much water must be added to 30 cm³ of a 0.2 mol/dm³ solution to dilute it to 0.02 mol/dm³?

We are given the initial volume ($V_1$), initial concentration ($C_1$), and desired final concentration ($C_2$). We need to find the final volume ($V_2$) and then the amount of water to be added.
The dilution formula is:

$$C_1V_1 = C_2V_2$$

Given:
$V_1 = 30 \text{ cm}^3$
$C_1 = 0.2 \text{ mol/dm}^3$
$C_2 = 0.02 \text{ mol/dm}^3$

$$(0.2 \text{ mol/dm}^3)(30 \text{ cm}^3) = (0.02 \text{ mol/dm}^3)V_2$$ $$6 \text{ mol} \cdot \text{cm}^3/\text{dm}^3 = (0.02 \text{ mol/dm}^3)V_2$$ $$V_2 = \frac{6 \text{ mol} \cdot \text{cm}^3/\text{dm}^3}{0.02 \text{ mol/dm}^3}$$ $$V_2 = 300 \text{ cm}^3$$

The amount of water to be added is the difference between the final volume and the initial volume.

$$\text{Water added} = V_2 - V_1$$ $$\text{Water added} = 300 \text{ cm}^3 - 30 \text{ cm}^3$$ $$\text{Water added} = 270 \text{ cm}^3$$

The amount of water that should be added is $\text{270 cm}^3$.