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 p pm
B. 10p pm
C. 20p pm
D. 16p pm

Answer

B

Explanation

`55.5 g CaCl_(2)=50 g"" CaCO_(3) implies 55.5 mg CaCl_(2)=50xx10^(-3)g CaCO_(3)`
`47.5 g MgCl_(2)=50g CaCO_(3) implies 47.5 mg MgCl_(2)=50xx10^(-3) g CaCO_(3)`
Total `CaCO_(3) =(50+50)xx10^(-3)=10^(-2) g""L^(-1)`
`p pm=(10^(-2))/(1000mL)xx10^(6)mL=10`