A sample of hard water contains `122 p p m` of `HCO_(3)^(ɵ)` ions,. What is the minimum weight of `CaO` required to remove ions completely from `1 kg` of such water sample?

Explanation

`[MwHCO_3^(ɵ)=61g,Mw(CaO)=56g]`
Temporary hardness is due to the `HCO_3^(ɵ)` of `Ca^(2+)` and `Mg^(2+)` and is removed by either boiling or by treating with CaO or `Ca(OH)_2` (Clark's method)
`underset(1 mmol)(Ca(HCO_3)_2)+underset(1 mmol)(CaO)tounderset(2 mmol)(2CaCO_3)+H_2O` ..(i)
`122ppm of HCO_3^(ɵ)-=122gHCO_3^(ɵ)` in `10^(6)mL or 10^(3)L`
`-=122 ml HCO_3^(ɵ)` in `1.0L`
`-=(122)/(61)=2 mmol` in `1.0 L`
`-=(2)/(2)` n-factor `=2`
`=1 m" mol of "Ca(HCO_3)_2` in `1.0L`
According to equation (i),
`1 m " mol of "Ca(HCO_3)_2-=1 m" mol of "CaO` in `1.0L`
`-=1xx56-=56" mg of " CaO`
`-=56 of CaO` in `1.0L`
`-=56 " mg of " CaO` in `1kg of H_2O`
`(1L of H_2O=1kg of H_2ObecausedH_2O=1)`