A sample of water contains 100mg/l of Ca2+ and 10 mg/l of Mg2+. Calculate the total hardness in mg/l as CaCO3?

Molar mass of Ca²⁺ = 40.08 g/mol
Molar mass of Mg²⁺ = 24.31 g/mol
Molar mass of CaCO₃ = 40.08 + 12.01 + 16.00×3 = 100.09 g/mol
(1 g/mol = 1 mg/mmol)

1 mole of Ca²⁺ is equivalent to the hardness of 1 mole of CaCO₃.
Also, 1 mole of Mg²⁺ is equivalent to the hardness of 1 mole of CaCO₃.

Consider 1 L of the water sample :
Number of milli-moles of Ca²⁺ = (100 mg) / (40.08 mg/mmol) = 2.50 mmol
Number of milli-moles of Mg²⁺ = (10 mg) / (24.31 mg/mmol) = 0.41 mmol
Total hardness as number of moles of CaCO₃ = (2.50 + 0.41) mmol = 2.91 mmol
Total hardness as mass of moles of CaCO₃ = (100.09 mg/mmol) × (2.91 mmol) = 291 mg

Total hardness is 291 mg/L as CaCO₃.