$$0.400$$ g of an organic compound (X) gave $$0.376$$ g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is (Given: Molar mass AgBr $$= 188$$ g mol$$^{-1}$$, Br $$= 80$$ g mol$$^{-1}$$)

In the Carius method, the organic bromine is converted to AgBr.

Moles of AgBr formed:

$$n_{AgBr} = \frac{0.376}{188} = 0.002 \text{ mol}$$

Since 1 mol AgBr contains 1 mol Br:

Mass of Br = $$0.002 \times 80 = 0.16$$ g

Percentage of bromine:

$$\% Br = \frac{0.16}{0.400} \times 100 = 40\%$$

The percentage of bromine is $$\mathbf{40}$$%.