$$10$$ L of hard water with temporary hardness of $$[Ca(HCO_3)_2]$$ requires $$0.56$$ g of lime. The reaction follows as:
Temporary hardness in terms of ppm of $$CaCO_3$$ is:
The correct option is D $$200$$ ppm
$$1$$ mole of calcium oxide corresponds to $$2$$ moles of calcium carbonate.
The molar mass of calcium oxide and calcium carbonate are $$56$$ g/mol and $$100$$ g/mol respectively.
$$0.56$$ g $$(0.01$$ mole$$)$$ of lime will correspond to $$2$$ g $$(0.02$$ mole$$)$$ of calcium carbonate.
Now, $$10$$ L of water corresponds to $$10000$$ ml.
Hence, $$2$$ g calcium carbonate in $$10000$$ ml of water corresponds to $$200$$ ppm.