The definition of parts per million:

1 g solute per 1,000,000 g solution

Now, divide both values by 1000 to get a new definition for ppm:

ppm = 0.001 g per 1,000 g solution

or:

ppm = 1 mg solute per 1 kg solution

Then, for an aqueous solution:

ppm = 1 mg solute per liter of solution

This last one works because the solution concentration is so low that we can assume the solution density to be 1.00 g/mL.

Also, it's this last modification of ppm (the mg/L one) that allows us to go to molarity (which has units of mol/L).

The best way to explain this is by doing some examples:

**Problem #1:** Convert 78 ppm of Ca^{2+} ions to mol/L

**Solution:**

1) By the last definition of ppm just above:

78 ppm = 78 mg Ca^{2+}/ L of solution = 0.078 g/L

2) Divide by the atomic weight for calcium ion:

0.078 g/L divided by 40.08 g/mol = 00019 mol/L

**Problem #2:** Calculate the molarity of a dye concentration given the molar mass is of the dye 327 g/mol and a dye concentration of 2 ppm.

**Solution:**

1) Convert ppm to a gram-based concentration:

2 ppm = 2 mg dye / L of solution

2a) Using 0.002 g/L, caculate the molarity:

0.002 g/L divided by 327 g/mol = 6.1 x 10^{-6}M

2b) Using 2 mg/L, calculate the molarity

2 mg/L divided by 327,000 mg/mol = 6.1 x 10^{-6}M

You might want to go back to problem #1 and try out 78 mg/L with the atomic weight of calcium ion expressed as mg/mol instead of g/mol.

**Problem #3:** A solution is labeled 2.89 ppm and is made with a solute that has molar mass equal to 522 g/mol. What is the molarity of the solution?

**Solution** (straight from Yahoo Answers):

1) Unless specified otherwise, ppm usually refers to ppm by weight:

2.89 ppm = 2.89 g per 1,000,000 g (or any other weight unit, I have just chosen g for convenience)

2) Now you need to know the density of the solvent to convert the volume to mass. If the solvent is water, we can assume the density at standard temperature and pressure is 1.0 g/mL. Therefore 1 litre (L) of water is 1,000 g. Therefore:

2.89 ppm = 2.89 g per 1,000 L = 0.00289 g per 1 L

(If you work out the mass per litre it makes working out the next steps a little easier, because the final units for molarity will be mol/L)

3) Now using the molar mass to work out the molarity (moles per litre):

Moles in one litre = mass/molar mass = 0.00289/522 = 5.5 x 10^{-6}mol

Therefore the molarity is 5.5 x 10^{-6} M

**Problem #4:** What is the ppm concentration of calcium ion in 0.010 M CaCO_{3}?

**Solution:**

1) Convert mol/L to to g/1000 g of solution:

0.010 mol/L times 40.08 g/mol = 0.4008 g/L0.4008 g/L = 0.4008 g / 1000 g of solution

Note that the solution density is assumed to be 1.00 g/mL. This is the common practice in ppm calculations.

2) Convert to ppm by multiplying numerator and denominator by 1000:

0.4008 g / 1000 g of solution times 1000/1000 = 400 g / 1,000,000 g of solutionThe answer is 400 ppm

Notice how the weight for calcium ion only is used. The weight of the carbonate, based on the wording of the question, is immaterial to the solving of the problem.

This focus on an ion (and not the entire formula of the substance) in ppm concentrations is common. Be aware!

**Problem #5:** Make a 20.0 ppm solution of HCl from a stock solution that is 0.500-molar.

**Solution:**

1) Let us assume we will make 1.00 L of solution. Since ppm = 1 mg solute per liter of solution:

we need 20.0 mg solute per liter of solution

2) Calculate the volume of 0.500 M solution that contains 20.0 mg HCl:

MV = g/molar mass(0.500 mol/L) (x) = 0.020 g / 36.46 g/mol

x = 0.001097 L

We need 1.10 mL of the stock solution diluted to 1.00 L to make a 20.0 ppm solution of HCl.

3) A slightly different form of the above question was answered thusly on Yahoo Answers:

The unit "ppm" is equivalent to "mg/L."Concentrated HCl is 12.3 moles/L or, since the molar mass of HCl is 36.5 g/L, 449 g HCl / L.

449 g/L = 449,000 mg/L. If you want 20 ppm, you need to make a 20/449,000 volume dilution, or 1/22,500.

The best way to do that is a serial dilution: first make a 1/100 dilution (1 mL conc. HCl / 100 mL) and then dilute THAT solution by 1/225. The result: 1/100 x 1/225 = 1/22,500.

You can cut the dilution process to one step if a lower concentration of HCl is available (e.g. 0.1 M).

**Problem #6:** In a geographical area at sea level measuring 56 km by 24 km by 1 km, the ozone concentration is 0.145 ppm. How many moles of ozone must be removed from the atmosphere to reach a concentration of 0.080 ppm, assuming a constant pressure of 1 atm and a constant temperature of 30 degrees Celsius.

**Solution:**

1) Calculate the liters (use cubic decimeters) in our volume:

(56 x 10^{5}dm) (24 x 10^{5}dm) (1 x 10^{5}dm) = 1.344 x 10^{18}dm^{3}

2) Calculate the ppm that must be removed:

0.145 - 0.080 = 0.065

3) Determine milligrams (then grams, then moles) of ozone to be removed from our volume:

1 ppm = 1 mg/LTherefore:

0.065 mg/L times 1.344 x 10

^{18}L = 8.736 x 10^{16}mg8.736 x 10

^{16}mg = 8.736 x 10^{13}g8.736 x 10

^{13}g divided by 47.997 g/mol = 1.82 x 10^{12}mol

**Problem #7:** 3100 mL of a river water sample took 9.30 mL of 0.01005 M Ag^{+} to titrate. Calculate the concentration of Cl¯ in ppm for the river water.

**Solution:**

1) Calculate moles of Ag^{+}:

(0.01005 mol/L) (0.00930 L) = 9.3465 x 10^{-5}mol of Ag^{+}

2) Since Ag^{+} and Cl¯ react in a 1:1 molar ratio:

9.3465 x 10^{-5}mol of Cl¯

3) How many mg of Cl¯ is this?

9.3465 x 10^{-5}mol x 35.453 g/mol = 3.31 x 10^{-3}g3.31 x 10

^{-3}g x (1000 mg/g) = 3.31 mg

4) ppm = mg/L

3.31 mg / 3.100 L = 1.07 ppm (to three sf)

**Problem #8:** A solution contains 6.0 x 10^{-6} mol Na_{2}SO_{4} in 250.0 mL. What is the concentration of sodium ion in ppm? Of sulfate ion?

**Solution:**

1) Convert to molarity:

6.0 x 10^{-6}mol / 0.250 L = 2.40 x 10^{-5}mol/L

2) Convert mol/L to grams of sodium ion per 1000 g of solution:

2.40 x 10^{-5}mol/L times 23.00 g/mol x 2 = 0.001104 g/L0.001104 g/L = 0.001104 g/1000 g of solution

The two is used because there are two sodium ions per formula unit of sodium sulfate.

For sulfate, the two would not be used and, instead of 23.00 g/mol, use the "formula weight" of the sulfate ion.

3) Convert to ppm by multiplying numerator and denominator by 1000:

0.001104 g/1000 g of solution times 1000/1000 = 1.104 g / 1,000,000 g of solutionThe answer (for sodium ion) is 1.1 ppm (to two sig figs).

The solution for sulfate is left to the reader.

**Problem #9:** The maximum concentration of O_{2} in seawater is 2.2 x 10^{-4} M at 25.0 °C. What is this concentration in ppm?

**Solution:**

1) Convert mol/L to grams of sodium ion per 1000 g of solution:

2.2 x 10^{-4}mol/L times 32.0 g/mol = 0.00704 g/L0.00704 g/L = 0.00704 g/1000 g of solution

2) Convert to ppm by multiplying numerator and denominator by 1000:

0.00704 g/1000 g of solution times 1000/1000 = 7.04 g / 1,000,000 g of solutionThe answer is 7.0 ppm (to two sig figs).