A solution of lead(II) chloride (M_r = 278.2) contains 1.08 g of PbCl₂ in 100 cm³ of solution - AQA - A-Level Chemistry - Question 26 - 2017 - Paper 3

Since one mole of PbCl₂ produces two moles of chloride ions (Cl⁻):
$\text{Number of moles of Cl}^- = 2 \times \text{Number of moles of PbCl₂} = 2 \times 0.00388 = 0.00776 \text{ mol}$

Concentration is defined as the number of moles per unit volume. Here, we have the total volume as 100 cm³, which is equivalent to 0.1 dm³. Therefore, the concentration of chloride ions is:
$\text{Concentration (mol dm}^{-3}\text{)} = \frac{\text{Number of moles of Cl}^{-}}{\text{volume (dm}^{3}\text{)}} = \frac{0.00776}{0.1} = 0.0776 \text{ mol dm}^{-3}$
This can be expressed as:
$7.76 \times 10^{-2} \text{ mol dm}^{-3}$