The correct option is B 71.43%
Given:CaCO3→CaO+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%