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 ` = (146 xx 10^4)/(10^6) = 1.46g`
`Mg(HCO_3)_2 +2Ca(OH)_2to 2CaCO_3 + Mg(OH)_2 +2H_2O`
146g of `Mg(HCO_3)_2 = 148g` of `Ca(OH)_2`
Weight of `Ca(OH)_2` required to precipitate 1.46 g of `Mg(HCO_3)_2` present in water sample ` = (148 xx 1.46)/(146)` = 1.48 g