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

Explanation

Degree of hardness of `H_(2)O` is 40 ppm, i.e., `10^(6)` g of water contain `CaCO_(3)`=40 g
Since 1 mole of `CaCO_(3)-=1 `mole of `MgSO_(4)`
`therefore 100 ` g of `CaCO_(3)-=120` g of `MgSO_(4)`
`therefore 10^(6)` g of water contain `MgSO_(4)=(40xx120)/(100)=48 g`
or `10^(3)` g of water contain `MgSO_(4)=(48xx10^(3))/(10^(6))xx10^(3) ` mg
or 1 kg of water will contain `MgSO_(4)=48` mg