One litre of a sample of hard water contains `5.55 mg " of "CaCl_(2)` and `4.75 mg" of "MgCl_(2)`. The total hardness in terms of ppm of `CaCO_3` is

Options

A. 5 ppm
B. 10 ppm
C. 20 ppm
D. 16 ppm

Answer

B

Explanation

`55.5 g Ca Cl _(2)-= 50 g Ca CO _(a) implies 55.5 mg Ca Cl _(2) -=50 xx 10 ^(-3) g CaCO _(3)`
`47.5 g Mg Cl _(2) -= 50 g Ca CO _(3) implies 47.5 mg M g Cl _(2)-= 50 xx 10 ^(-3) g Ca CO _(3)`
Total `Ca CO _(3) = ( 50 + 50) xx 10 ^(-3) = 10 ^(-2) g L ^(-1)`
`p p m = ( 10 ^(-2))/( 1000 m L) xx 10 ^(6) m L = 10`