Using the relationship M1V1 = M2V2 where M1 and M2 are the molar concentrations (mol/L or mmol/ml) and V1 and V2 are the volumes of the solutions, we can arrive at the following answer for the given problem:
15.0M (L of stock solution) = 2.35M (0.25L) *all volumes were converted to liters.
L of stock solution = (2.35*0.25)/15.0
Therefore, 0.0392L or 39.17 ml of stock solution is needed.
