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//Chapter-17,Example 2,Page 369
clc();
close();
m1= 136 // mass of FeSO4
m2 = 100 //mass of CaCO3
//for 100 ppm hardness FeSO4 required per 10^6 parts of water is 136 parts
//for 200 ppm hardness
amt= m1*200/m2
printf("the amount of FeSO4 required is = %.f mg/l",amt)
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