The concentration of lead ion in a water sample is 5ppb. Calculate the amount of lead ion in 2000g of this water sample?

10 μg or 1 x 10⁻⁵ g

Explanation:

The concentration of lead ion is given in parts by billion (ppb), which means that there are 5 micrograms (μg) of lead in 1 liter of water sample:

5 ppb = 5 μg lead ion/L

If we assume that the density of the water sample is equal to 1 g/ml (density of water at 25°C), we can say that 2000 g of sample is equal to 2000 mL:

sample volume = 2000 g x 1 mL/g = 2000 mL x 1 L/1000 mL = 2 L

Thus, the amount of lead ion in 2000 g (or 2 L) of water sample is:

mass of lead ion = 5 μg lead/L x 2 L = 10 μg x 1 g/10⁶μg = 1 x 10⁻⁵ g

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