The mass of $\mathrm{Ca}(\mathrm{OH}{)}_2 ,74\times {10}^{-x}gm$, required for 10 liter of water remove temporary hardness of 100 ppm due to $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ ? $[\text{ At wt }\mathrm{Ca}=40,\mathrm{O}=16,\mathrm{H}=1]$. What is the x value

10⁶ gr of water contain 100 gr of CaCO₃ Lit of 10⁴ gr water contain
$\begin{array}{l}\frac{100}{106}\times {10}^4=1\mathrm{gr}\text{ of }{\mathrm{CaCO}}_3\\ 100\mathrm{gr}\text{ of }{\mathrm{CaCO}}_3=162\mathrm{gr}\text{ of }\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2\\ 1\mathrm{gr}{\mathrm{CaCO}}_3=1.62\mathrm{gr}\text{ of }\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2\left(\text{ or }\right)0.01\text{ Moles }\\ \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2+\mathrm{Ca}(\mathrm{OH}{)}_2⟶2{\mathrm{CaCO}}_3(\perp )+2{\mathrm{H}}_2\mathrm{O}\end{array}$
Moles of $\mathrm{Ca}(\mathrm{OH}{)}_2=0.01$
Weight  of $\mathrm{Ca}(\mathrm{OH}{)}_2=0.74\mathrm{gr}$