Mass of \(A g B r=141 \mathrm{mg}=0.141 \mathrm{g}\)

\(1\; mole\) of \(A g B r=1\; g\) atom of \(B r\)

\(188 g\) of \(A g B r=80 g\) of \(B r\)

\(188 g\) of \(A g B r\) contain bromine \(=80 g\)

\(0.141 g\) of \(A g B r\) contain bromine \(=\frac{80}{188} \times 0.141\)

This much amount of bromine present in \(0.250 \mathrm{g}\) of organic compound

\(\because \%\) of bromine \(=\frac{80}{188} \times \frac{0.414}{0.250} \times 100=24 \%\)