100kg of a water sample contains 6g of magnesium sulphate, 10.2g of calcium sulphate and 16.2 g of calcium bicarbonate only. Calculate permanent hardness, temporary hardness and total hardness.

Explanation

`100 Kg (10^5g)` of given water sample contains 6 g of `MgSO_4, 10.2 g CaSO_4` and `16.2g Ca(HCO_3)_2` .
Weight of `MgSO_4` in `10^6`g water `= (6)/(10^5) xx 10^6= 60g`
120 g of `MgSO_4` = 100 g of `CaCO_3`
60 g of `MgSO_4 = 60/120 xx 100 = 50 g` of `CaCO_3`
Degree of hardness due to `MgSO_4` = 50ppm
weight of `CaSO_4` in `10^6g` of water = `10.2 xx 10^6//10^5 = 102g`
136 g of `CaSO_4` = 100 g of `CaCO_3`
102 g of `CaSO_4 = 102 xx 100//136 = 75` ppm
Degree of hardness due to `CaSO_4 = 75` ppm
Weight of `Ca(HCO_3)_2` present in `10^6` gm of water ` = 16.2 xx 10^6//10^5 = 162 g `
162 g of `Ca(HCO_3)_2 -= 100 g `of `CaCO_3`
Degree of hardness due to `Ca(HCO_3)_2 = 100` ppm
Temporary hardness of water sample = 100ppm
Permanent hardness of water sample = 50 + 75 = 125 ppm
Total hardness of water sample = 100+125=225 ppm