Describe how you would prepare 200ml of a 0.25 solution of a naoh deom a 3m stock solution

This is a question of dilution. We are given with concentrated solution of NaOH (3M), from which we have to prepare a dilute solution (0.25M) of volume 200mL. We will use following formula M1V1=M2V2 M1 = Initial conc = 3M V1 = volume required M2 = final xonc=0.25M V2 = desired volume of final solution V1 = 0.25X200/3 = 16.67 mL So we will take 16.67 ML of concentrated 3M NaOH and will add water to make total volume to 200mL.