In: Chemistry
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
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