Calculte the degree of hardness of a sample of water containing `6 mg` of `MgSO_(4)` per kg of water.

Answer

5

Explanation

`because 1 g-"mole"` or `120 g MgSO_(4) "has" =1 g-"mole"` or `100g CaCO_(3)`
`therefore 6xx10^(-3)g MgSO_(4) "has" =(100xx6xx10^(-3))/(120)`
`=5xx10^(-3)gCaCO_(3)`
Thus `1000 g` of water contains `MgSO_(4)` equivalent to `5xx10^(-3)gCaCO_(3)`
`therefore 10^(6)g` water contains `=(5xx10^(-3))/(1000)xx10^(6)`
`5 g "of" CaCO_(3)`