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+//Chapter 14: Water Treatment
+//Problem: 10
+clc;
+
+//Initialisation of Variables
+v1 = 50.   //in ml for hardwater
+v2 = 15    //in ml for EDTA
+m = 0.01   //in M for EDTA
+
+//Solution
+M = v2 * m / v1
+N = M * 2
+S = N * 50 * 1000
+
+mprintf("Molarity of hardness is :%.3f M\n", M)
+mprintf(" Normality of hardness is :%.3f N\n", N)
+mprintf(" Strength of hardness is :%d ppm", S)