Question
A sample of water is found to contain 0.059 ppm of Pb2+ ions. Calculate the mass of lead ions per liter of this solution (assume the density of the water solution is 1.0 g/mL)
1.2E-2 g/L Pb2+
5.9E-8 g/L Pb2+
5.9E-5 g/L Pb2+
1.2E-5 g/L Pb2+
2.9E-10 g/L Pb2+
1.2E-2 g/L Pb2+
5.9E-8 g/L Pb2+
5.9E-5 g/L Pb2+
1.2E-5 g/L Pb2+
2.9E-10 g/L Pb2+
Answers
DrBob222
A good conversion factor to remember is
1 ppm = 1 mg/L (1 ppb = 1 ug/L)
So 0.059 ppm = 0.059 mg/L
Then convert mg to g and you have it.
1 ppm = 1 mg/L (1 ppb = 1 ug/L)
So 0.059 ppm = 0.059 mg/L
Then convert mg to g and you have it.