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clear;
clc;
printf("\t Example 7.1\n");
//let x be the weight of water in the quantity of solution needed
c=.498; //solute content afetr crystallisation
W1=111; //molecular weight of CaCl2
W2=219; //molecular weight of CaCl2.6H2O
M1=(108/W2)*100; //water present in 100kg of CaCl2.6H2O
M2=(W1/W2)*100; //CaCl2 present in 100kg of CaCl2.6H20
//t=M2+c*x; //total weight entering the solubility
//x+49.3; total water solubility used
//s*(x+49.3)/100 //total Cacl2 after solubility
x=poly([0],'x'); //calc. x the weight of crystal
t=roots((M2+c*x)-(x+49.3)*.819);
printf("\nthe weight of water in the quantity of solution needed :%f kg",t);
h=(c)*t; //weight of CaCl2 corresponding to weight water
tw=t+h; // total weight of the solution
printf("\nthe total weight of the solution is :%f kg",tw);
//end
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