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 of ppm of `CaCO_(3)` is

Options

A. 5 ppm
B. 10 ppm
C. 20 ppm
D. None of these

Answer

B

Explanation

`55.5 g CaCl_(2) -= 50 g CaCO_(3)`
`55.5 mg CaCl_(2) -= 50 xx 10^(-3) CaCo_(3)`
`47.5 g MgCl_(2) -= 50 g CaCO_(3)`
`47.5 mg Mg Cl_(2) -= 50 xx 10^(-3) g CaCO_(3)`
Total `CaCO_(3) = (50 + 50) xx 10^(-2) = 10^(-2) g L^(-1)`
`ppm = (10^(-2))/(1000 mL) xx 10^(6) mL = 10`