The concentration of Cd in lake sediment digestate is 3 µg/L. If 5 g of dry...
The concentration of Cd in lake sediment digestate is 3 µg/L. If 5 g of dry sediment was used, and the total digestate volume is 25 mL, what is the Cd concentration in sediment?
Wouldn't the mass of Cd be the 5g?
Homework Answers
3 µg/L means 3 µg Cd remain in 1 L lake water.
5 g of dry sediment was used, and the total digestate volume is 25 mL = 0.025 L
thus
mass of Cd in 0.025 L lake water = 3 µg/L * 0.025 L = 0.075 µg = 0.075 * 10^-6 g
and
Cd concentration in sediment = 0.075 * 10^-6 g * 10^6 / 5 = 0.015 ppm (answer)
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