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diff --git a/905/CH1/EX1.2/1_2.sce b/905/CH1/EX1.2/1_2.sce
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+clear;
+clc;
+
+// Illustration 1.2
+// Page: 7
+
+printf('Illustration 1.2 - Page: 7\n\n');
+
+//*****Data*****
+// Component  a-KNO3   b-H20
+T = 293; // [K]
+s_eqm = 24; // [percent by weight, %]
+row = 1162; // [density of saturated solution, kg/cubic m]
+//*****//
+
+printf('Illustration 1.2 (a)- Page: 7\n\n');
+// Solution (a)
+
+// Basis: 100 kg of fresh wash solution
+m_a = (s_eqm/100)*100; // [kg]
+m_b = 100 - m_a; // [kg]
+M_a = 101.10; // [gram/mole]
+M_b = 18.02; // [gram.mole]
+// Therefore moles of component 'a' and 'b' are
+n_a = m_a/M_a; // [kmole]
+n_b = m_b/M_b; // [kmole]
+
+m_total = 100; // [basis, kg]
+n_total = n_a+n_b; // [kmole]
+// Average molecular weight
+M_avg = m_total/n_total; // [kg/kmole]
+// Total molar density of fresh solution
+C = row/M_avg; // [kmole/cubic m]
+printf("Total molar density of fresh solution is %f kmole/cubic m\n\n",C);
+
+printf('Illustration 1.2 (b)- Page: 8\n\n');
+// Solution (b)
+
+// mole fractions of components 'a' and 'b'
+x_a = n_a/n_total;
+x_b = n_b/n_total;
+printf("Mole fraction of KNO3 and H2O is %f %f",x_a,x_b); 
+\ No newline at end of file