Question

In: Chemistry

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

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

Solutions

Expert Solution

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


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