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diff --git a/905/CH3/EX3.3/3_3.sce b/905/CH3/EX3.3/3_3.sce
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+clear;
+clc;
+
+// Illustration 3.3
+// Page: 162
+
+printf('Illustration 3.3 -  Page: 162\n\n');
+
+// solution
+//*****Data*****//
+// a-ammonia   b-air   c-water
+T = 300; // [K]
+P = 101.3; // [kPa]
+R = 8.314; // [cubic m.Pa/mole.K]
+V_b = 15; // [cubic m]
+m_a = 10; // [kg]
+m_c = 45; // [kg]
+M_a = 17; // [molecular mass of ammonia, gram/mole]
+M_c = 18; // [molecular mass of water, gram/mole]
+//*****//
+
+n_b = V_b*P/(R*T); // [kmole]
+n_a = m_a/M_a; // [kmole]
+n_c = m_c/M_c; // [kmole]
+
+// L_a as the number of kmol of ammonia in the liquid phase when equilibrium is achieved
+// And n_a-L_a kmol of ammonia will remain in the gas phase
+// x_a = L_a/(n_c+L_a)                     (1)
+// y_a = (n_a-L_a)/(n_b+n_a-L_a)           (2)
+// gamma = 0.156+0.622*x_a*(5.765*x_a-1)   (3)  for x_a <= 0.3 
+// y_a = 10.51*gamma*x_a;                  (4)
+// Equations (1),(2),(3), and (4) are solved simultaneously
+deff('[y] = f12(L_a)','y = ((n_a-L_a)/(n_b+n_a-L_a))-(10.51*(0.156+(0.622*(L_a/(n_c+L_a))*(5.765*(L_a/(n_c+L_a))-1)))*(L_a/(n_c+L_a)))');
+L_a = fsolve(0.3,f12); // [kmole]
+
+x_a = L_a/(n_c+L_a);
+y_a = (n_a-L_a)/(n_b+n_a-L_a);
+gammma_a = 0.156+0.622*x_a*(5.765*x_a-1);
+
+printf("At equilibrium the ammonia content of the liquid phase will be %f\n\n",x_a);
+printf("At equilibrium the ammonia content of the gas phase will be %f\n\n",y_a);
+printf("The amount of ammonia absorbed by the water will be %f kmole\n\n",L_a);  
+\ No newline at end of file