Question

Solution of 100 mL water contains $0.73{\text{ g}}$ of ${\text{Mg}}{\left( {{\text{HC}}{{\text{O}}_3}} \right)_2}$ and $0.81{\text{ g}}$ of ${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_3}} \right)_2}$ . Calculate the hardness in terms of ppm of ${\text{CaC}}{{\text{O}}_3}$ .
A.${10^2}$ ppm
B.${10^4}$ ppm
C.$5 \times {10^3}$ ppm
D.${10^3}$ ppm

Answer 1

To Calculate hardness we basically need to calculate the molarity. Using given mass and molar mass we will calculate molarity. To convert it into the ppm we multiply it by a factor of ${10^6}$ .

Complete step by step answer:
The molar mass of ${\text{Mg}}{\left( {{\text{HC}}{{\text{O}}_3}} \right)_2}$ is 146 and the molar mass of ${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_3}} \right)_2}$ is 162. Number of moles is the ratio of given mass by the molar mass and hence the number of moles will be:
Number of moles of ${\text{Mg}}{\left( {{\text{HC}}{{\text{O}}_3}} \right)_2}$ will be $\dfrac{{0.73}}{{146}}$ and the number of moles of ${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_3}} \right)_2}$ will be $\dfrac{{0.81}}{{162}}$ .
The total volume is 100 and hence the hardness in ppm will be:
$\dfrac{{\dfrac{{0.73}}{{146}} + \dfrac{{0.81}}{{162}}}}{{100}} \times 100 \times {10^6} = {10^4}$

Hence, the correct option is B.

Additional information:
Parts per million is a term to define the concentration. It indicates the parts of the given substance that is present in ten lakh or 1 million parts. This unit is usually used for defining the substances whose concentration is low for example the impurities in air is also calculated in the ppm. The term ppm is used to do that concentration because observable and hence can be studied otherwise the concentration will be very less.

Note:
There are two types of hardness of water: temporary hardness and permanent hardness. Temporary hardness is caused by the carbonates and bicarbonates of calcium and magnesium present in water whereas the permanent hardness is caused by the sulphates and chlorides present in water. The temporary hardness is easily removed by boiling water. The carbonates are released as gases but chemical methods are needed to remove the permanent hardness o f water.