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

1a.

Goal: make a 5 mM standard of methylene blue

MB molar mass= 319.85 g/mol

only

Question

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 equation since we’re looking for volume and we know the volume and concentration of one. I tried getting the concentration of the other part by taking my 0.79 grams from part a and dividing by the molar mass, getting .00247 mol. I then divided by 0.025 L because that’s the volume we need. I ended up with 0.098M and completed the problem by doing (.002M x 0.025L) =(v2 x 0.98 M), but I got a really weird number at the end and I don’t think I did it right.

for 2a I did m1v1=m2v2 for all and got the following respectively: 0.235L; 0.18L; 0.12L; 0.065L; .02L

I don’t know how to make serial dilutions

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