0.400 g of an organic compound $(\mathrm{X})$ gave 0.376 g of AgBr in Carius method for estimation of bromine. $\%$ of bromine in the compound $(X)$ is $\_\_\_\_$ .
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$\begin{aligned} & \text { mole of } \mathrm{AgBr}=\frac{0.376}{188} \\ & \text { mole of } \mathrm{Br}^{-}=\text {mole of } \mathrm{AgBr}=\frac{0.376}{188} \\ & \text { mass of } \mathrm{Br}^{-}=\frac{0.376}{188} \times 80 \\ & \% \text { of } \mathrm{Br}^{-}=\frac{0.376 \times 80}{188 \times 0.4} \times 100=40 \%\end{aligned}$
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