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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+
+clear;
+clc;
+printf("\t Example 7.5\n");
+
+p1=.3; //percentage of the solute in the solution
+w1=1000; //weight of the solution taken
+w2=142; //molecular weight of Na2SO4.
+M1=(w2/(180+w2)); //solute (Na2SO4) present in the Na2CO3.10H2O solution
+s1=40.8; //solubility of Na2SO4 at 30 degree per 100 gm of water
+s2=9.0; //solubility of Na2SO4 at 10 degree per 100 gm of water
+//percent weight of solute in Na2SO4.10H2O= 144/322
+//let x be the weight of crystal formed
+x=poly([0],'x'); //calc. x the weight of crystal
+t=roots((w1*40.8/140.8)-(.442*x+(w1-x)*(s2/(100+s2))));
+printf("\n the weight of crystal formed after crystallisation :%f kg",t);
+
+//end \ No newline at end of file