In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr. The mass percentage of bromine in the compound is ______.
(Atomic mass, Ag = 108, Br = 80 g \(\mathrm{mol^{-1}}\))
(Atomic mass, Ag = 108, Br = 80 g \(\mathrm{mol^{-1}}\))
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Correct Answer: 50
Solution and Explanation
Mass of Br = \(\dfrac{80}{188} \times 1.88\) = 0.80 g.
% of Br = \(\dfrac{0.80}{1.6} \times 100\) = 50.
Answer: 50
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