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Question

$$10$$ L of hard water with temporary hardness of  $$[Ca(HCO_3)_2]$$ requires $$0.56$$ g of lime. The reaction follows as:
$$Ca(HCO_3)_2+CaO\longrightarrow 2CaCO_3+H_2O$$.
Temporary hardness in terms of ppm of $$CaCO_3$$ is:


A
56 ppm
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B
12 ppm
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C
100 ppm
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D
200 ppm
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Solution

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.

Chemistry

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