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0.4g of an organic compound gave 0.188g of silver bromide by a halogen estimation method. The percentage of bromine in the compound is (at.wt of Ag=108, Br=80)

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Detailed Solution

% $Br = \frac{80}{188} \times \frac{Mass of AgBr \times 100}{Mass of compound}$

$= \frac{80}{188} \times \frac{0.188 \times 100}{0.4} = \frac{8}{0.4} = 20 %$

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