# How is 1ppm=1mg/L ?

1. Sep 23, 2015

### CivilSigma

1. The problem statement, all variables and given/known data

I know for a fact that $1ppm = 1\frac{mg}{L}$ , but frankly I don't understand how that is derived. If someone can show me how, I would really appreciate it!

2. Relevant equations

3. The attempt at a solution

2. Sep 23, 2015

### marcus

It would only be true for fluids with standard density where a liter weighs 1000 grams.

In that case 1 gram dissolved in a liter would be one part per thousand by mass.

And one milligram in a liter would be one thousandth of that, namely one part per million.

3. Sep 24, 2015

### Staff: Mentor

I wouldn't call them "standard density" (or at least, I have never seen 1 g/mL referred to as such).

1 g/mL is a good approximation of the density of most water solutions. Seawater has a density around 1.025 g/mL, and it already contains plenty of dissolved salts. As most solutions we deal with are moderately concentrated water solutions, this approximation (1 g/mL) is quite universal.

Even for concentrated solutions (or even liquids that don't contain water) we rarely get really far from 1 g/mL. Pure sulfuric acid gets up to 1.84 g/mL, concentrated ammonia down to 0.86 g/mL, so we are never far from the approximation 1 ppm = 1 mg/L.

4. Sep 24, 2015

### CivilSigma

So is that just the definition of ppm and there is no derivation?

5. Sep 24, 2015

### Staff: Mentor

1 ppm by definition means a ratio - of 1 something per million of something (but the definition doesn't say ratio of what). It can be 1 gram per 1000000 grams (which is equivalent of 1 mg/1 kg), it can be 1 mL/1 cubic meter, it can be even 1 atom per 106 atoms. Assuming it is weight/weight ratio and we are talking about water solutions of density close to 1 g/mL it can be easily shown that 1 ppm is equivalent to 1 mg/1 L.

In most cases, when there is additional information ppm means weight/weight - but it is not guaranteed. I believe I have read about someone running into serious problems after misreading gas contamination levels listed in ppm as w/w instead of v/v (or vice versa).

6. Sep 24, 2015

### CivilSigma

Oh I see.

So for $\frac{mg/L} = \frac{g}{10^3 \cdot 10^3 g} =\frac{g}{10^6 g} =ppm$ assuming the density is: $1000=\frac{Mass}{1 L}$.

7. Sep 24, 2015

### Staff: Mentor

Your LaTeX is broken to the point it is hard to make sense out of it, but in general it is not difficult to convert units and show why ppm and mg/L are equivalent.

8. Sep 24, 2015

### CivilSigma

Ok, as an example , is this how you would convert the following into ppm

0.002 mg/g = 0.002 mg/ L ( given 1000g/L = mass/Volume) = 0.002 ppm ?

9. Sep 25, 2015

### Staff: Mentor

No, because 1 g is not equivalent to 1 liter. 0.002 mg/L is 1 ppb, not 1 ppm.

If anything, 0.002 mg/g is equivalent to 0.002 mg/1 mL.

10. Sep 25, 2015

### CivilSigma

Are we using the density of water to determine the relation ships between ml and grams , L and m^3 etc? Because I don't see how they are equal.

But for 0.002 mg/g = 0. 000 002 g/g = 0. 000 002 = x ppm/10^6 : x= 2ppm ?

11. Sep 25, 2015

### Staff: Mentor

All the time, from the very first post. 1 L of water weighs 1 kg (or 1000 g), that's almost the definition of the density.

12. Sep 25, 2015

### CivilSigma

Oh okay, so how about m^3?

13. Sep 26, 2015

### Staff: Mentor

You don't know how to calculate mass of a cubic meter, knowing the density is 1 g/mL?

14. Sep 26, 2015

### CivilSigma

I don't see it....

$\frac{1g}{1ml} = \frac{unknown \, mass}{1m^3} \implies unknown \, mass = \frac{1g}{10^{-3} \cdot 1000g} \cdot m^3 = 1 ?$

15. Sep 27, 2015

### Staff: Mentor

I am afraid nothing is correct here. While the first part correctly refers to the fact density is a ratio of mass and volume it incorrectly assumes it is numerically identical no matter what kind of units are used. Then you are throwing in a bunch of conversion factors and scramble units.

By definition density is d=m/V, you know the density (1 g/mL) and the volume (1 m3). You need to convert the units so that you don't mix mL and m3. Then you just solve d=m/V for m and plug the numbers.