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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)
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