`10 L` of hard water required `0.56 g` of lime `(CaO)` for removing hardness. Hence, temporary hardness in ppm (part per million `10^(6)`) of `CaCO_(3)` is:

Options

A. `100`
B. `200`
C. `10`
D. `20`

Answer

B

Explanation

Temporary hardness is due to `HCO_(3)^(ɵ)` of `Ca^(2+)` and `Mg^(2+)`
`Ca(HCO_(3))_(2)tounderset(56 g)(CaO)tounderset((2xx100)g)(2CaCO_(3))+H_(2)O`
`0.56 g CaO-=2g CaCO_(3)` in `10 L H_(2)O`
`=2g CaCO_(3)` in `10^(4) mL H_(2)O` ltbRgt `=20 g CaCO_(3)` in `10^(6) mL H_(2)O`
`=20 p p m of CaCO_(3)`