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//Chapter-17,Example 5,Page 370
clc();
close();
v= 50000 //volume of water
m1 = 84 //mass of MgCO3
m2 = 100 //mass of CaCO3
m3 = 95 //mass of MgCl2
m4 = 111 //mass of CaCl2
amnt_1 = 144 //amount of MgCO3 in ppm
amnt_2 = 25 //amount of CaCO3 in ppm
amnt_3 = 95 //amount of MgCl2 in ppm
amnt_4 = 111 //amount of CaCl2 in ppm
lime = (74/100)*[2*(amnt_1*100/m1)+(amnt_2*100/m2)+(amnt_3*100/m3)]*v
printf("the lime required is = %.3f mg",lime)
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