1L of a sample of hard water contains 1 mg `"CaCl"_(2)` and 1 mg `"MgCl"_(2)`. Estimate the degree of hardness of this sample of water.

Explanation

Molecular mass of `"CaCl"_(2)` = 111
Now 111g of `"CaCl"_(2) equiv` 100g of `"CaCO"_(3)`
`therefore ` 1 mg `CaCl_(2) equiv (100 xx 0.001)/(111)`g of `CaCO_(3)`
`= 9 xx 10^(-4)`g of `"CaCO"_(3)` = 0.9mg of `CaCO_(3)`
Molecular mass of `MgCl_(2)` = 95
Now, 95g of `MgCl_(2) equiv 100g " CaCO"_(3)`
`therefore` 1 mg of `"MgCl"_(2) equiv (100)/(95)` mg of ` "CaCO"_(3)` = 1.05 mg of `CaCO_(3)`
Equivalent amount of `"CaCO"_(3)` corresponding to `"CaCl"_(2)`
and `MgCl_(2)` present in 1 L or `10^(3)` g of hard water
= (0.90 + 1.05) mg = 1.95 mg [ 1L water = `10^(3)`g = `10^(6)` mg water]
Mass of equivalent amount of `CaCO_(3)` corresponding to `CaCl_(2) and MgCl_(2)` present in `10^(6)` mg of water is 1.95 mg. Hence, degree of hardness of the given sample is 1.95 ppm.