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+clear;
+clc;
+
+// Example: 9.2
+// Page: 338
+
+printf("Example: 9.2 - Page: 338\n\n");
+
+// Solution
+
+//*****Data******//
+Vol = 2;// [Volume of the mixture, cubic m/mol]
+y1 = 0.4;// [mole fraction of alcohol, cubic m/mol]
+V1_bar = 38.3*10^(-6);// [partial molar volume of alcohol, cubic m/mol]
+V2_bar = 17.2*10^(-6);// [partial molar volume of water, cubic m/mol]
+V1 = 39.21*10^(-6);// [molar volume of alcohol, cubic m/mol]
+V2 = 18*10^(-6);// [molar volume of water, cubic m/mol]
+//*************//
+
+// From Eqn. 9.9:
+V = y1*V1_bar + (1 - y1)*V2_bar;// [molar volume of the solution]
+n = Vol/V;// [number of moles of solution]
+n1 = y1*n;// [number of moles of alcohol required]
+n2 = (1 - y1)*n;// [number of moles of water required]
+V_alcohol = V1*n1;// [Volume of alcohol required, cubic m]
+V_water = V2*n2;// [Volume of water required, cubic m]
+printf("Volume of alcohol required is %.3f cubic m while volume of water required is %.3f cubic m\n",V_alcohol,V_water); \ No newline at end of file