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# Given,$10L$ of hard water required $0.56g$ of lime ($CaO$) for removing hardness. Hence the temporary hardness in ppm of $CaC{O_3}$ is?A) $100$B) $200$C) $10$D) $20$

Last updated date: 16th Jul 2024
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Hint: We have to remember that the units part per million is ${10^6}$ mL of any solution. We also remember that the hardness of water can be measured in parts per million calcium carbonate, which is the amount of calcium carbonate in gms present in ${10^6}$ ml of water. Hardness can be of two types: permanent and temporary.

$Ca{\left( {HC{O_3}} \right)_2} + CaO \to {\text{ }}2CaC{O_3} + {H_2}O$
We must have to know that the molar mass of lime is $56g$ and on the other hand molar mass of calcium carbonate is $100g$ since two moles of calcium carbonate is required thus $200g$. So from here we can conclude that $0.56g$ of calcium oxide i.e. lime requires $2g$ of calcium carbonate i.e. $CaC{O_3}$.
$0.56g$ of $CaO$= $2g$ of $CaC{O_3}$ in $10L$ of water
= $2g$ $CaC{O_3}$ in ${10^4}$ mL of ${H_2}O$
= $200g$ $CaC{O_3}$ in ${10^6}$ ml of ${H_2}O$