How many milliliters of a 5.15 M stock solution are needed to prepare 375 mL of
a solution having 7.50 mg of methanol per mL of solution?
7.5 mg MeOH/mL x 1000 mL = 7500 mg/L or 7.5 g/L.
mols MeOH = grams/molar mass and M = mols/L.
Then use the dilution formula of
mL1 x M1 = mL2 x M2