Question

A sample of an industrial waste water is analyzed and found to contain 15.2 ppb Cu2+. How many grams of copper could be recovered from 1.23×10^3 kg of this waste water?

Answer 1

Firstly we will express ppb as mass per mass:

`1.0\text{ }ppb=(1.0\mu g)/(1kg)`

By using this definition we can determine the copper in the waste water sample:

`15.2ppb\text{ }Cu^(2+)=\frac{15.2\mu g\text{ }Cu^(2+)}{1kg\text{ }wastewater}`

If 15.2 micrograms of copper is in 1 Kg of wastewater then in 1.23x10^3 kg of waste water we will have:

`\begin{gathered} 15.2\mu g=1kg \\ x\mu g=1.23*10^3kg \\ x=(15.2\mu g*(1.23*10^3kg))/(1kg) \\ x=18,696\mu g \end{gathered}`

Answer: 18,696 micrograms of copper would be in 1.23x10^3kg of wastewater.