Calculate the mass of ammonium chloride (NH₄Cl) needed to prepare 200 cm³ of a 0.5 mol dm⁻³ solution.

The atomic masses are given as N = 14.0, H = 1.0, Cl = 35.5.

$$\text{Molar mass of NH}_4\text{Cl} = (1 \times 14.0) + (4 \times 1.0) + (1 \times 35.5)$$ $$\text{Molar mass of NH}_4\text{Cl} = 14.0 + 4.0 + 35.5 = 53.5 \text{ g/mol}$$

The concentration is 0.5 mol dm⁻³.

$$\text{Moles} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles} = 0.5 \text{ mol/dm}^3 \times 0.2 \text{ dm}^3 = 0.1 \text{ mol}$$

$$\text{Mass} = \text{Moles} \times \text{Molar mass}$$ $$\text{Mass} = 0.1 \text{ mol} \times 53.5 \text{ g/mol} = 5.35 \text{ g}$$

The mass of ammonium chloride crystals that will be needed is $\boxed{\text{5.35 g}}$.