The degree of hardness of a sample of water is 40 ppm. If the hardness is only due to the presence of `"MgSO"_(4)` , then determine the amount of `"MgSO"_(4)` in 1 kg of that water.

Explanation

The hardness of the sample of water is 40 ppm.
Therefore, `10^(6)` g of that sample contains 40 g `CaCO_(3)`.
`therefore` 1 kg or `10^(3)` g of water contains `(40 xx 10^(3))/(10^(6))` = 0.04 g `CaCO_(3)`
Now, 1 mol `"CaCO"_(3) equiv` 1 mol `"MgSO"_(4)`
or, 100 g `"CaCO"_(3) equiv` 120 g `"MgSO"_(4)`
[ molecular mass of `"CaCO"_(3)` = 100 and `"MgSO"_(4)` = 120]
or, 0.04 g `"CaCO"_(3) equiv (120 xx 0.04)/(100)`g or, 0.048g `"MgSO"_(4)`
`therefore` Hence, the amount of `"MgSO"_(4)` present per kg of that water is 0.048 g or 48 mg.