Question

Given,$10L$ of hard water required $0.56$ of lime (\[CaO\]) for removing hardness. Hence, the temporary hardness in ppm (part per million) of $CaC{O_3}$ is:
(a)100
(b)200
(c)10
(d)20

Answer 1

Hint: Volume is the third way to measure the amount of matter, after item count and mass. Volume varies greatly depending on the density of the substances. Avogadro explained that the volumes of all gases can be easily determined.

Complete answer:
\[CaO\] is mixed with water to produce $Ca{(OH)_2}$ and them this is added to $Ca{(HC{O_3})_2}$ to precipitate $CaC{O_3}$ (Clarke’s process of removal of temporary hardness).
It is given that, $10L$ of hard water require $0.56$g of \[CaO\],
 The molecular mass of \[CaO\] is=$40 + 16 = 56g$
The balanced equation of the Clarke’s process is:
$Ca{(HC{O_3})_2} + CaO \to 2CaC{O_3} + {H_2}O$
$1$ mole of \[CaO\] gives $2$ moles of $CaC{O_3}$, so we can calculate the number moles of \[CaO\]in $10L$
No of moles= $\dfrac{\text{Given mass}}{\text{Molecular mass}} = \dfrac{{0.56}}{{56}} = 0.01$
So, number of moles of $CaC{O_3}$ formed in $10L$ is= $0.01 \times 2 = 0.02$ moles of $CaC{O_3}$
Molecular mass of $CaC{O_3}$=$40 + 12 + 3(16) = 100g$
Mass of $CaC{O_3}$ = No of moles $ \times $ Molecular mass = $0.02 \times 100 = 2g$$ = 2000mg$
Mass of $CaC{O_3}$ in $1$L= $\dfrac{{2000}}{{10}} = 200mg$
$1ml$ of water, mass= $1g$
$1000ml$ of water mass= $1000g = {10^3}g$
= ${10^3} \times {10^3}mg$
$ = {10^6}mg$
Amount of $CaC{O_3}$ present in per ${10^6}$ part of ${H_2}O$= $200$
And hence option b is the correct answer.

Note:
In Clark’s reaction, hard water is treated with slaked lime. Calcium hydroxide is Clark’s reagent. It removes the hardness of water by converting bicarbonates into carbonates. This method involves the addition of slaked lime to water either in solid or in liquid form. This results in the conversion of soluble bicarbonates into soluble carbonates.