Using 18% by weight ammonium hydroxide to make 16ppm aqueous solution

[Frankenstein19]

TL;DR Summary: Used M1V1=M2V2 but am unsure if formula applies

I feel like I could use M1V1=M2V2 but the 18% by weight ammonium hydroxide is confusing me so much.

Density of ammonium hydroxide 18% is 0.926 g/cm3

18%V1=16mg/L*200L

(18mg ammonium/100 mg of total solution) * V1 = 3,200mg ammonium (aq)

(18mg ammonium/100 mg of total solution) * V1 = 3,200mg ammonium (aq)

V1 = 3,200mg ammonium / 18mg ammonium * 100mg of total solution


V1 = 17,777 mg of solution × 1g/1000mg × cm3/0.926g

V1 = 1.91 cm3

Is this correct?

 

[Borek]

You even didn't wrote what is the volume you are trying to prepare - is it 200 L?

No, M1V1=M2V2 is not the right approach. You should not operate with volumes, but with masses.

18 ppm is typically w/w - that is, you want your solution to contain 18*10^-6 ammonia by mass.

What is mass of the 200 L of the solution? (with no other data I would use just the density of pure water).

18 ppm of that - how many grams of ammonia required?

How many grams of the 18% solution contain required mass of ammonia?

And finally - convert to the volume.

 

[symbolipoint]

The desired finished concentration is 16 mg. of ammonium hydroxide per 1000 ml. of solution. Helpful?

Let me jump ahead...
The amount of GRAMS of the base you'd have if you took v ml of the base solution(which is 18%) is ((0.18 grams ammnmhydxd)/(1 gram stock))((0.926 grams stock)/(1 ml. stock))v .


Some simplifications and keeping in mind you want to prepare solution resulting in 16 mg. per 1000 ml,

((0.18)(0.926)v)/1000=16