In an estimation of bromine by the 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 mol^–1)

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Solution

Solution: According to Carious method,
% of Bromine (Br) =(weight of AgBr × atomic weight of Br ×100) / (molecular weight of AgBr ×weight of organic bromide)
=(1.88×80×100) / (188×1.6)
= 15040/300.8
= 50%
So, in the estimation of bromine by the Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr. The mass percentage of bromine in the compound is 50%.

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In an estimation of bromine by the 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 mol–1)

In an estimation of bromine by the 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 mol^–1)