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What is the total hardness of sample of water has the following impurities in mg/I. Ca(HCO3)2 = 162 CaCI2 = 22.2 MgCI2 = 95 NaCI = 20
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Given:

Ca(HCO3)2=162CaCl2=22.2MgCl2=95NaCl=20

Solution:

Equivalents of Calcium Carbonate=Mass of hardness producing substance×50Chemical equivalent of hardness producing substance

1.

Ca(HCO3)2=162×50162.11=49.966 mg/L

2.

CaCl2=22.2×50110.98=10 mg/L

3.

MgCl2=95×5095.21=49.889 mg/L

4.

NaCl=20×5040=25 mg/L

 

Total Hardness=49.97+10+49.89+25=134.86 mg/L

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