(Solved): We'll be making up standard solutions of \( \mathrm{Fe}^{2+} \) from a stock solution that is 40 pp ...

We'll be making up standard solutions of \( \mathrm{Fe}^{2+} \) from a stock solution that is 40 ppm (parts per million, or 1 mg solute per 1 Liter of solution) in \( \mathrm{Fe}^{2+} \). Let's review how to calculate the ppm and molarity of the diluted standard solution for known #6, and then you can do the calculation for unknowns 1-5. Known #6 uses \( 5 \mathrm{~mL} \) of \( 40 \mathrm{ppm} \mathrm{Fe}^{2+} \) stock, and ends up diluting it to a new total volume of \( 50 \mathrm{~mL} \). You can use \( M_{1} V_{1}=M_{2} V_{2} \) to calculate the new ppm for known #6: \( (40 . \mathrm{ppm})(5.0 \mathrm{~mL})=\mathrm{M}_{2}(50 . \mathrm{mL}) \) \( M_{2}=4.0 p p m \) To calculate the molarity of this solution, we covert \( \frac{4.0 \mathrm{mg} \mathrm{Fe}}{\mathrm{L} \text { of solution }}+\frac{\text { nol of } \mathrm{Fe}^{2+}}{\sqrt{\text { of solution }}} \) In a similar way, calculate the ppm and molarity of each of the other knowns, Known \( 1-5 \)