`10 L` of hard water requires `0.28 g` of line `(CaO)` for removing hardness. Calculate the temporary hardness in ppm of `CaCO_(3)`.

Explanation

Temporary hardness is due to `HCO_(3)^(ɵ)` of `Ca^(2+)` and `Mg^(2+)`
`Ca(HCO_(3))_(2) + underset(56 g)(CaO) rarr underset(2 xx 100 g)(2CaCO_(3) + H_(2)O)`
`56 g CaO = 200 g CaCO_(3)` in `10 L of H_(2) O`
`0.28 g CaO = (200 xx 0.28)/(56)`
`1 g CaCO` in `10 L of H_(2)O`
`= 1 g CaCO_(3)` in `10 xx 1000 mL of H_(2) O`
`= 1 g CaCO_(3)` in `10^(4) mL of H_(2) O`
`100 g CaCO_(3)` in `10^(6) mL of H_(2) O = 100 ppm`
Hence, temporary hardness of `CaCO_(3) = 100 ppm`