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authorprashantsinalkar2018-02-03 10:59:42 +0530
committerprashantsinalkar2018-02-03 10:59:42 +0530
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parentf35ea80659b6a49d1bb2ce1d7d002583f3f40947 (diff)
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-//Chapter-17,Example 4,Page 370
-clc();
-close();
-
-m1 = 146 //mass of Mg(HCO3)2
-
-m2 = 162 //mass of Ca(HCO3)2
-
-m3 = 111 //mass of CaCl2
-
-m4 = 120 //mass of MgSO4
-
-m5 = 136 //mass of CaSO4
-
-amnt_1 = 12.5 //amount of Mg(HCO3)2 in ppm
-
-amnt_2 = 10.5 //amount of Ca(HCO3)2 in ppm
-
-amnt_3 = 8.2 //amount of CaCl2 in ppm
-
-amnt_4 = 2.6 //amount of MgSO4 in ppm
-
-amnt_5 = 7.5 //amount of CaSO4 in ppm
-
-temp_hard= (amnt_1*100/m1)+(amnt_2*100/m2)
-
-perm_hard= (amnt_3*100/m3)+(amnt_4*100/m4)+(amnt_5*100/m5)
-
-total= temp_hard +perm_hard
-
-printf("the temporary hardness is = %.3f mg/l",temp_hard)
-
-printf("\n the permanent hardness is = %.3f mg/l",perm_hard)
-
-printf("\n the total hardness is = %.3f mg/l",total)
-
-v= 100 //volume of sample
-
-v_EDTA = total*v/1000 //volume of EDTA
-
-printf("\n the volume of M/100 EDTA required is = %.3f ml",v_EDTA)