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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.

Answer» <html><body><p></p>Solution :`100 Kg (10^5g)`of given water <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> contains 6 g of `MgSO_4, 10.2 g CaSO_4` and `16.2g <a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a>(HCO_3)_2` .<br/> Weight of `MgSO_4`in `10^6`g water `= (6)/(10^5) xx 10^6= 60g`<br/>120 g of `MgSO_4` = 100 g of `CaCO_3` <br/>60 g of `MgSO_4 =60/120 xx 100 = 50 g` of `CaCO_3`<br/>Degree of <a href="https://interviewquestions.tuteehub.com/tag/hardness-14971" style="font-weight:bold;" target="_blank" title="Click to know more about HARDNESS">HARDNESS</a> due to `MgSO_4` = 50ppm<br/>weight of `CaSO_4` in `10^6g` of water = `10.2 xx 10^6//10^5 = 102g`<br/>136 g of `CaSO_4` = 100 g of `CaCO_3` <br/>102 g of `CaSO_4 = 102 xx 100//136 = <a href="https://interviewquestions.tuteehub.com/tag/75-334971" style="font-weight:bold;" target="_blank" title="Click to know more about 75">75</a>` ppm<br/>Degree of hardness due to `CaSO_4 = 75` ppm<br/>Weight of `Ca(HCO_3)_2` present in `10^6` gm of water ` = 16.2 xx 10^6//10^5 = 162 g `<br/>162 g of `Ca(HCO_3)_2 -= 100 g `of `CaCO_3` <br/>Degree of hardness due to `Ca(HCO_3)_2 = 100` ppm<br/> Temporary hardness of water sample = 100ppm<br/>Permanent hardness of water sample = 50 + 75 = 125 ppm<br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> hardness of water sample = 100+125=225 ppm</body></html>


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