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Question

200 g of a calcium carbonate sample decomposes on heating to give carbon dioxide and 80 g of calcium oxide. What will be the percentage purity of calcium carbonate in the sample?
(Given:CaCO3CaO+CO2)

A
58.14%
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B
71.43%
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C
84.33%
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D
28.89%
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Solution

The correct option is B 71.43%
Given:CaCO3CaO+CO2
1 mole of CaO is produced by the decomposition of 1 mole of CaCO3
i.e. 56 g of CaO is produced by 100 g of CaCO3
80 g will be produced by 10056×80=142.85 g of CaCO3
% purity =mass of pure compound in sampletotal mass of impure sample×100=142.85200×100
=71.43%

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