Answer:
Mass of organic compound, W =
0.15 g
Mass of\[AgBr\],
\[{{W}_{1}}=0.12g\]
\[\underset{188g}{\mathop{AgBr}}\,\equiv
\underset{80g}{\mathop{Br}}\,\]
Percentage of
bromine = \[\frac{80\times {{W}_{1}}}{188\times W}\times 100\]
\[=\frac{80\times 0.12\times 100}{188\times 0.15}=34.04%\]
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