Answer:
[tex]\large\boxed{0.019\, \mu \text{mol/L}}[/tex]
Explanation:
[tex]\text{Concentration } = \dfrac{\text{moles}}{\text{litres}}\\\\\text{c} = \dfrac{n}{V}[/tex]
1. Convert millilitres to litres
[tex]\text{V = 350. mL} \times \dfrac{\text{1 L}}{\text{1000 mL}} = \text{0.3500 L}[/tex]
2. Calculate the concentration
[tex]c = \dfrac{0.0067\, \mu\text{mol}}{\text{0.3500 L}} = \mathbf{0.019 \, \mu}\textbf{mol/L}\\\\\text{The concentration of mercury(II) iodide is $\large\boxed{\mathbf{0.019\, \mu} \textbf{mol/L}}$}.[/tex]
