Question

The degree of hardness of a given sample of hard water is `40 p p m`. If the entire hardness is due to `MgSO_(4)`, how much of `MgSO_(4)` is present per `kg` of water?

Answer 1

Degree of hardness of water `=40 p p m` i.e., `10^(6) g` of water contains `CaCO_(3)=40 g` Since `1` mole of `CaCO_(3)=1 mol e of MgSO_(4)`
`100 g of CaCO_(3) =120 g MgSO_(4)`
`:. 10^(6) g` of water contains `=(40xx120)/100=48 g MgSO_(4)`
`10^(3) g` water contain =`(48xx10^(3))/(10^(6))xx10^(3)MgSO_(4)`
Or `1 kg` of water contains =`48 mg of MgSO_(4)`