posted by ChEm .
A stock solution was prepared by dissolving exactly 0.4000 g of pure ASA (180.16 g/mol)
in 10.00 mL of NaOH and heating the solution to a gentle boil. After cooling to room
temperature, the solution was poured into a 250-mL volumetric flask and diluted to the mark
with DI water. Before coming to lab, calculate the concentration of ASA in the flask.
I got my molarity as 0.00888M:
however am I suppose to take into account the 10.0ml of NaOH and if so how would I do that?