Question

The Annual Drinking Water Report states 0.46 ppm chromium and 9.87 ppb lead as inorganic contaminants.

a) How many micrograms of each contaminant are contained in 1 milliliter of water?

b) Give the concentration of each contaminant in mol/L

Answer 1

1 ppm = 0.001 ml/l; 1 ml/l = 1000 ppm

1 ng/ml = 1 ppb; 1 ppb = 1 ng/ml

1 ug/L = 0.001 ppm; 1 ppm = 1000 ug/L

a)

given that

0.46 ppm chromium and 9.87 ppb lead

0.46 ppm chromium * 1000 ug/L =460 ug/L

                                           1 ppm

460 ug/L Cr is present in 1.0 L or 1000 ml

So

1 milliliter of water will contains: 460 ug/ 1000 mL

= 0.46 micrograms Cr in 1 milliliter of water

9.87 ppb lead

1ppm = 1000ppb

Thus, 9.87 ppb lead

= 0.00987 ppm

0.00987 ppm lead * 1000 ug/L =9.87 ug/L

                                           1 ppm

9.87 ug/L Pb is present in 1.0 L or 1000 ml

So

1 milliliter of water will contains: 9.87 ug/ 1000 mL

= 0.00987 micrograms Pb in 1 milliliter of water

b)

here; 460 ug/L or 0.00046 g/ L Cr is present in 1.0 L

because 1.0 ug = 10^-6 g

now calculate the number of moles = amount in g / molar mass

fro Cr = 0.00046 g/ L /51.9961 g/ mole

= 8.85M or moles / L Cr

And; 9.87 ug/L or 9.87*10^-6 g/ L Pb is present in 1.0 L

now calculate the number of moles = amount in g / molar mass

fro Pb = 9.87*10^-6 g/ L /207.2 g/ mole

= 4.76 M or moles / L Pb