Answer:
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
