Calculate the mass of ammonium chloride crystals needed to make 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.

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

Given volume = $200 \text{ cm}^3$.
Since $1 \text{ dm}^3 = 1000 \text{ cm}^3$:

$$\text{Volume} = \frac{200}{1000} \text{ dm}^3 = 0.2 \text{ dm}^3$$

The concentration is $0.5 \text{ mol dm}^{-3}$ and the volume is $0.2 \text{ dm}^3$.

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

$$\text{Mass} = \text{Moles} \times \text{Molar mass}$$ $$\text{Mass} = 0.1 \text{ mol} \times 53.5 \text{ g mol}^{-1} = 5.35 \text{ g}$$

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