One litre of a sample of hard water contains 1 mg of `CaCl_(2)` and 1 mg of `MgCl_(2)`. Find the total hardness of water in terms of parts of `CaCO_(3)` per `10^(6)` parts of water by mass.

Options

A. 1.05 ppm
B. 1.95 ppm
C. 0.95 ppm
D. 95 ppm

Answer

B

Explanation

Mol. Mass of `CaCl_(2)=111`,
Mol. Mass of `MgCl_(2)=95`
111 grams of `CaCl_(2)=100` g of `CaCO_(3)`
`therefore` 1 mg of `CaCl_(2)=100/111 times 1` mg of `CaCO_(3)`
= 0.9 mg of `CaCO_(3)`
95 g of `MgCl_(2)=100g " of "CaCO_(3)`
`therefore` 1 mg of `MgCl_(2)=100/95 times 1` mg of `CaCO_(3)`
= 1.05 mg of `CaCO_(3)`
thus, one litre of hard water contains (0.9+1.05)
= 1.95 mg of `CaCO_(3)`
one litre water `=10^(3)g=10^(6)mg`
Degree of hardness = 1.95 ppm