[ahmed1]

if you have a tank V=783 m3, and 90% of tank is filled with fresh cooling water. you need minimum 600 ppm of nitrite to be present with water in the tank. you do so by adding the corrosion inhibitor (sodium nitrite).
Question
how much in kg of sodium nitrite is required for every 100 ppm of nitrite?

solution
783*.9=705 m3 is the volume of water=705000 L
600 ppm= 600 mg/L=0.006 kg/L (simple conversion)
0.006*705000=4230 kg of nitrite. MW of nitrite=46 kg/kmol, so 4230/46=92 kmol
by formula
NaNo2+H2O=NaOH+H(+)No2(-)

92 kmol of nitrite needs 92 kmol of sodium nitrite
92kmol*69kg/kmol (MW of sodium nitrite)=6345 kg this amount is to be added to maintain 600 ppm nitrite in water
so for only 100 ppm. 6345/6

I need your help in checking the steps if right or wrong