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+
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.
To calculate the mass of lead (Pb2+) ions per liter of the solution, we need to convert the concentration of Pb2+ ions from parts per million (ppm) to grams per liter (g/L).
First, let's convert the given concentration of Pb2+ ions from ppm to g/L:
1 ppm = 1 g in 1 million g
= 1 g in 1,000,000 mL (since density = 1.0 g/mL)
= 1 g in 1,000 L
So, the concentration of Pb2+ ions in g/L can be calculated as:
0.059 ppm Pb2+ ions * (1 g/1,000,000 g) * (1,000 mL/1 L)
= 0.059 * 1/1,000,000 * 1,000
= 0.059 * 1/1,000
= 5.9E-5 g/L Pb2+ ions
Therefore, the correct option is:
5.9E-5 g/L Pb2+