Question

$10$ L of hard water with temporary hardness of $\left[Ca\right(HC{O}_3{)}_2]$ requires $0.56$ g of lime. The reaction follows as:
$Ca(HC{O}_3{)}_2+CaO⟶2CaC{O}_3+H_{2}O$.
Temporary hardness in terms of ppm of $CaC{O}_3$ is:

• A. $56$ ppm
• B. $\frac{1}{2}$ ppm
• C. $100$ ppm
• D. $200$ ppm

Answer 1

The correct option is D $200$ ppm

$1$ mole of calcium oxide corresponds to $2$ moles of calcium carbonate.
The molar mass of calcium oxide and calcium carbonate are $56$ g/mol and $100$ g/mol respectively.
$0.56$ g $(0.01$ mole$)$ of lime will correspond to $2$ g $(0.02$ mole$)$ of calcium carbonate.
Now, $10$ L of water corresponds to $10000$ ml.
Hence, $2$ g calcium carbonate in $10000$ ml of water corresponds to $200$ ppm.