Answer:
Here's what I get
Explanation:
Assume you want to use 0.5 mL of a stock solution with a concentration of 0.2 mol·L⁻¹.
You want to prepare a 0.002 mol·L⁻¹ solution.
1. Moles of Fe(NO₃)₃
[tex]\text{Moles of Fe(NO$_{3}$})_{3} = \text{0.5 mL stock} \times \dfrac{\text{0.2 mmol Fe(NO$_{3})$}_{3}}{\text{1 mL stock}}\\\\= \text{0.1 mmol Fe(NO$_{3})$}_{3}[/tex]
2. Volume of dilute solution
[tex]V = \text{0.1 mmoL Fe(NO$_{3})$}_{3}\times \dfrac{\text{1 mL solution}}{\text{0.002 mmoL Fe(NO$_{3})$}_{3}}= \textbf{50 mL solution}[/tex]
