Question

Hardness of water is measured in terms of ppm (parts per million) of $CaC{O}_3$. It is the amount (in g) of $CaC{O}_3$ present in ${10}^{6}g{H}_{2}O$. In a sample of water, $10$ L required $0.56$ g of $CaO$ to the remove temporary hardness of $HC{O}_3^{-}$.
$Ca(HC{O}_3{)}_2+CaO⟶2CaC{O}_3+H_{2}O$
Temporary hardness is :

• A. $200$ ppm $CaC{O}_3$
• B. $100$ ppm $CaC{O}_3$
• C. $50$ ppm $CaC{O}_3$
• D. $25$ ppm $CaC{O}_3$

Answer 1

The correct option is A $200$ ppm $CaC{O}_3$

$Ca(HC{O}_3{)}_2+CaO⟶2CaC{O}_3+H_{2}O$

$10L{H}_{2}O(={10}^{4}mL)hasCaC{O}_3=2g$

${10}^{6}mL\left(ppm\right)hasCaC{O}_3=200g$

Thus, $200ppmCaC{O}_3$ is the temporary hardness.