The amount of water that should be added to 500 ml of 0.5 N solution of NaOH to give a concentration of 10 mg / ml is

N1=0.5N→10mg per mLN2=10×10−3g40×1×1000=0.25NV1=500ml,V2=?N1V1=N2V2;0.5×500=0.25×V2V2=1000mL final volume water added −1000−500=500mL.