• $\mathrm {1 \ litre \ of \ NaCl \ solution \ = 50 \ g\ of\ NaCl }$
• $\mathrm {\therefore 200 \ litres \ of \ NaCl \ solution \ contains \ = 50\times 200 = 10000\ g \ of\ NaCl}$
• $\mathrm {But, \ 58.5g \ of\ NaCl\ = 50g \ of \ CaCO_3}$
• $\mathrm{10000g\ NaCl\ in \ terms \ of \ CaCO_3 \ equivalent }=\dfrac{10000\times 50}{58.5} = 8547.01\ g$ of CaCO3
• $\mathrm {This \ will\ be \ the \ hardness\ that\ can \ be \ removed\ by \ zeolite\ softner}.$

$\mathrm {8547\times1000 \ mg \ of \ CaCO_3 }$

$\mathrm {8547000ppm \ CaCO_3.}$

• $\mathrm {As\ 50ppm \ CaCO_3 \ equivalent\ is\ the \ hardness\ in \ 1 \ litre.}$
• $\mathrm {So, 8547000 \ CaCO_3 \ equivalent \ hardness \ in = \dfrac{8547000}{50} = 170940 \ Litres\ of\ water}$