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$$0.400$$ 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$$^{-1}$$)
Correct Answer: 40
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 $$\mathbf{40}$$%.
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