Ashamed but please help me figure out this chemistry

[ReefLab]

I dissolved 30g of KNO3 in 420mL of RODI Water. How many mL of solution do I add to my 400L tank to raise NO3 by 5ppm?
I got 21.9mL but the online calculators are not confirming that..

(wrong) Math: 61.3% NO3 by mass gives 18.40g NO3 into 420mL. That is a concentration of 43.81g/L. To raise 400L 5ppm, 2g of KNO3 is needed.. 43.81/2=21.9? Pretty sure something is wrong in that last step..

 

[JimWelsh]

You are very close. You need to invert that last part. What you need is 2/43.81 = 0.0457 liters, or 45.7 mL.

 

[Shirak]

18.4g in 420ml is .0438g/ml xml/2g=1ml/.0438g solve x 2/.0438 = 45.6ml
check 45.6 x .0438 =1.997g

 

[ReefLab]

You are very close. You need to invert that last part. What you need is 2/43.81 = 0.0457 liters, or 45.7 mL.

Ahh there it is. Thanks a ton!

18.4g in 420ml is .0438g/ml xml/2g=1ml/.0438g solve x 2/.0438 = 45.6ml
check 45.6 x .0438 =1.997g

Love proportions. Thanks for laying it out for me.

 

[JimWelsh]

To raise 400L 5ppm, 2g of KNO3 is needed.

For the record, that's 2g of NO3, not KNO3. You already did that conversion with the "61.3%" factor.