Question

$10\, L$ of hard water required $0.56 \,g$ of lime $( CaO )$ for removing hardness
$Ca \left( HCO _{3}\right)_{2}+ CaO -2 CaCO _{3}+ H _{2} O$
Thus, temporary hardness in terms of $ppm$ of $CaCO _{3}$ is

• A. 10
• B. 20
• C. 100
• D. 200

Answer 1

Answer: (D)

$\begin{matrix}Ca \left( HCO _{3}\right)_{2}+ CaO&\rightarrow&2 CaCO _{3}+ H _{2} O\\ 1\,mol\qquad&&2\,mol\\ 56\,g&&200\,g\\ 0.56\,g&&2.00\,g\end{matrix}$

Thus, $10\, L$ water $\left(=10^{4} \,g \right)$ water has hardness $=2\, g$

$\therefore 10^{6}\, g ( ppm )$ water has hardness $=\frac{2 \times 10^{6}}{10^{4}}=200$