Question

The Odessa water report states 0.44 ppm nitrate and 4.1 ppb lead as inorganic contaminants.
a) how many micrograms of each contaminant are contained in 1 mL of water?
b) give the concentration of each contaminant in mol/L

Answer 1

Definition of parts per million in water is

miligrams / Liter

Nitrate:

we have 0.44 ppm in 1 liter, we want to know how many miligrams are in 1 ml

1 ml = 0.001 L

miligrams of nitrate = 0.44 * 0.001 = 0.000 44 miligrams

1 miligram has 1000 micrograms so, we have 0.44 micrograms of nitrate.

To convert to mol / L

0.44 ppm = 0.44 miligrams / Liter = 0.000 44 grams / Liter

we just need to divide by the molar mass of nitrate 62

0.000 44 grams / Liter * 1 / 62 g/gmol = 7.0967 x 10^-6 mol / L

For the lead

Definition of ppb is

ppb = micrograms / Liter

1 ml = 0.001 Liter

a) micrograms = 0.001 * 4.1 = 0.0041

To change to mol / L

4.1 ppb = 4.1 micrograms / Liter

1 microgram = 1 x 10^-6 grams

so we have

4.1 x 10^-6 grams / Liter

Now we only need to divide by the molar mass of lead 207.2 g/gmol

4.1 x 10^-6 grams / Liter * 1 / 207.2 g/gmol = 1.9788 x 10^-8 mol / L