Question

The maximum contaminant level, of copper ions in drinking water is 1.3 ppm. If your drinking water contains .424 ppm Cu, how many ions of copper(I) will be in a sample of drinking water?

Answer 1

0.424 ppm Cu means that 1 L of drinking water contains 0.424 mg Cu (since 1 ppm = 1 mg/L)

Atomic mass of Cu = 63.546 g/mol.

Mol(s) Cu corresponding to 0.424 mg Cu = (0.424 mg)/(63.546 g/mol)

= (0.424 mg)*(1 g/1000 mg)/(63.546 g/mol)

= ((4.24*10^-4 g)/(63.546 g/mol)

= 6.6723*10^-6 mol.

It is known that 1 mol of a substance contains 6.02*10²³ units (atoms or ions) of the substance.

Therefore, number of Cu(I) ions corresponding to 6.6723*10^-6 mol copper =

(6.6723*10^-6 mol)*(6.02*10²³ ions)/(1 mol)

= 4.0167*10¹⁸ ions

≈ 4.02*10¹⁸ ions (ans, correct to 3 sig. figs).