1a. Goal: make a 5 mM standard of methylene blue MB molar mass= 319.85 g/molonly

1a.

Goal: make a 5 mM standard of methylene blue

MB molar mass= 319.85 g/mol

only

glassware available is 500mL flask

how much solid MB do we need?

1b.

you need a standard of 2 mM MB (molar mass 319.85 g/mol)

you need: 25 mL of new solution

How much stock solution from problem 1 will you need for a dilution

2a. stock of 50 mM

you need 5 250mL solutions at the following concentrations

47 mM, 36 mM, 24 mM, 13 mM, and 4 mM

2b. Now, we need to make a serial dilution

Calculate how you would make each solution (47 mM, 36 mM, 24 mM, 13 mM, and 4 mM)

I'm stuck on these four problems. For 1a, I converted 5mM to .005M and multiplied by the 500 mL (.5 L) and got .0025 mol. From there, I multiplied the .0025 by the molar mass of 319.85 g/mol and got 0.79 grams for the first part. For part b, I'm guessing we have to use the M1V1=M2V2