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authorprashantsinalkar2018-02-03 10:59:42 +0530
committerprashantsinalkar2018-02-03 10:59:42 +0530
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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)