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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)
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