Michele needs to make 1 L solutions of 10 mM solutions of ammonium acetate frequently. Instead of
Question:
Michele needs to make 1 L solutions of 10 mM solutions of ammonium acetate frequently. Instead of weighing out the ammonium acetate solid each time, she decides to make a 100 mL concentration stock solution of ammonium acetate and pipet 1 mL of it into 1 L to obtain the 10 mM desired concentration. How much ammonium acetate must she weigh out to make the concentrated stock she will need?
2. Michele's plan from #1 works so well, she decides to also make a concentration ammonium carbonate solution to use in the same way. How much must she weigh out to make another 100 mL of the new concentrated stock?
3. Michele weighs 0.01025 grams of analyte and dissolves this into 10 mL total volume. If this solution is then diluted 10.00 µL into 1000 µL what is the final concentration in ppm (w/v)?
4. If Michele's instrument uses 10.0 µL of the solution from #3 above, how many µg of analyte does this involve?
5. Michele has an analyte with a molecular weight of 600.02 g/mol. If she wants to make a 1.000 mg/mL solution, how much of 1 M NaOH stock solution must she pipet to have the same number of moles of NaOH as the analyte?