In Carius method of estimation of halogen, $$0.45 \text{ g}$$ of an organic compound gave $$0.36 \text{ g}$$ of $$AgBr$$. Find out the percentage of bromine in the compound. (Molar masses: $$AgBr = 188 \text{ g mol}^{-1}; Br = 80 \text{ g mol}^{-1}$$)

In the Carius method for estimation of bromine, the organic compound is heated with fuming $$HNO_3$$ in the presence of $$AgNO_3$$. The bromine in the compound is converted to $$AgBr$$.

Given:

Mass of organic compound = $$0.45 \text{ g}$$

Mass of $$AgBr$$ formed = $$0.36 \text{ g}$$

Molar mass of $$AgBr = 188 \text{ g mol}^{-1}$$

Atomic mass of $$Br = 80 \text{ g mol}^{-1}$$

$$ \text{Moles of } AgBr = \frac{0.36}{188} $$

Since 1 mole of $$AgBr$$ contains 1 mole of $$Br$$:

$$ \text{Moles of } Br = \frac{0.36}{188} $$

$$ \text{Mass of } Br = \frac{0.36}{188} \times 80 = \frac{28.8}{188} = 0.15319 \text{ g} $$

$$ \% \text{ of } Br = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 $$

$$ \% \text{ of } Br = \frac{0.15319}{0.45} \times 100 = 34.04\% $$

The correct answer is Option A: $$34.04\%$$.