10 L of hard water required 5.6 g of lime for removing hardness. Hence temporary hardness in ppm of

Question

10 L of hard water required 5.6 g of lime for removing hardness. Hence temporary hardness in ppm of $\mathrm{CaCO}_3$ is:

Professor PC Thomas Classes · Answer

(a) 
$ \begin{aligned} & \mathrm{Ca}\left(\mathrm{HCO}_3ight)_2+\mathrm{CaO} \longrightarrow \mathrm{CaCO}_3+\mathrm{H}_2 \mathrm{O} \ & 162 \mathrm{~g}-56 \mathrm{~g} \ & 16.2 \mathrm{~g}-5.6 \mathrm{~g} \ & 162 \mathrm{~g} \mathrm{Ca}\left(\mathrm{HCO}_3ight)_2=100 \mathrm{~g} \mathrm{CaCO}_3 \ & 16.2 \mathrm{~g} \mathrm{Ca}\left(\mathrm{HCO}_3ight)_2=10 \mathrm{~g} \mathrm{CaCO}_3 \ & 	ext { Hardness }=\frac{	ext { Mass of CaCO }}{	ext { Mass of water }} 	imes 10^6 \ & \quad=\frac{10}{10^4} 	imes 10^6=1000 \mathrm{ppm} \end{aligned} $