Question

$10$ L of hard water required $0.56$ g of lime $\left(CaO\right)$ for removing hardness. Hence, temporary hardness in ppm of $CaC{O}_3$ is :

• A. $100$
• B. $200$
• C. $10$
• D. $20$

Answer 1

The correct option is B $200$

The bicarbonates of calcium and magnesium result in the temporary hardness of water.

The balanced chemical reaction for the reaction between calcium bicarbonate and magnesium bicarbonate is as given below:

$Ca(HC{O}_3{)}_2+CaO\to 2CaC{O}_3+H_{2}O$
56 g ------ 200g

0.56 g ------ 2g

$10L({10}^4$ mL$)$ of hard water contains $2g$ of calcium carbonate.

Hence, ${10}^{6}ppm$ of hard water will contain $200g$ of calcium carbonate.