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Question

10 L of hard water with temporary hardness of [Ca(HCO3)2] requires 0.56 g of lime. The reaction follows as:
Ca(HCO3)2+CaO2CaCO3+H2O.
Temporary hardness in terms of ppm of CaCO3 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.

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