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.  

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