Degree of hardness of a water sample is 100ppm. If that sample contains hardness causing salt `Mg(HCO_3)_2`, calculate weight of `Ca(OH)_2` required for the treatment of 10kg of water sample.

Explanation

Degree of hardness is 100ppm. Thus `10^(6)` g of hard water contains 100g of `CaCO_(3)` which is equivalent to 146g of `Mg(HCO_(3))_(2)`
Weight of `Mg(HCO_(3))_(2)` present in 10kg of water `=(146xx10^(4))/(10^(6))=1.46g`
`Mg (HCO_(3))_(2)+2Ca(OH)_(2)to2CaCO_(3)+Mg(OH)_(2)+2H_(2)O`
146g of Mg`(148zxx1.46)/(146)=1.48g`