diff options
Diffstat (limited to '821/CH6/EX6.1/6_1.sce')
| -rwxr-xr-x | 821/CH6/EX6.1/6_1.sce | 12 |
1 files changed, 12 insertions, 0 deletions
diff --git a/821/CH6/EX6.1/6_1.sce b/821/CH6/EX6.1/6_1.sce new file mode 100755 index 000000000..a256603fe --- /dev/null +++ b/821/CH6/EX6.1/6_1.sce @@ -0,0 +1,12 @@ +MW=249.6;//molecular weight of CuSO4.5H2O in grams//
+w=0.3120;//weight of CuSO4.5H2O in grams//
+V=0.25;//volume of the solution in litres//
+printf('From Equation (a) 2 mol of CuSO4.5H2O liberates 1 mol of I2,i.e. 2 equivalents.\nHence the equlivalent weight of CuSO4.5H2O=mol.wt/1.');
+printf('\nFrom equation (b) the equivalent weight of CuSO4.5H2O is mol.wt/2 since 1mol of CuSO4.5H2O reacts with 2 mol of OH-,i.e 2 equivalents.');
+W=w/V;//weight of CuSO4.5H2O in one litre solution in grams//
+printf('\nWeight of CuSO4.5H2O in a litre of the solution=W=%fgrams.',W);
+Na=W/MW;//Normality of the solution for (a)//
+printf('\nNormality of the solution for (a)=Na=%f',Na);
+Nb=W*2/MW;//Normality of the solution for (b)//
+printf('\nNormality of the solution for (b)=Nb=%f',Nb);
+printf('\nIn the first case 1ml of the solution contains 5*10^-3equivalents or 5 equivalents of CuSO4.5H2O,\nand in the second case 1ml of the solution will contain 10m eq of CuSO4.5H2O.');
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