Question

If 0.495 g of material is completely dissolved in deionized water to make a 100 mL...

If 0.495 g of material is completely dissolved in deionized water to make a 100 mL solution and that solution is found to be 0.865 ppm in lead; what was the concentration of lead in the original material?

Homework Answers

Answer #1

Ans: 175 ppm

(Mass of lead/Mass of solution) x 106 = 0.865

(Mass of lead/Mass of solution) = 0.865 x 10-6

Mass of solution = mass of material + mass of water

Mass of solution = 0.495 g + 100 g = 100.495 g   (because density of water = 1 g/cc)

{Also, it is assumed that 100 mL of water is added to make 100 mL solution}

Mass of lead in the solution = (0.865 x 10-6) x (100.495) = 8.693 x 10-5 g

Mass of lead/Mass of material = (8.693 x 10-5)/0.495 = 0.000175

(Mass of lead/Mass of material) x 106 = 0.000175 x 106 = 175 ppm

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