In Carius method of estimation of halogen, $$0.25\,g$$  of an organic compound gave  $$0.15\,g$$  of silver bromide $$(AgBr).$$ The percentage of Bromine in the organic compound is $$\underline{\hspace{2cm}}\times 10^{-1}$$ (Nearest integer). }[ Given: Molar mass of Ag is 108 and Br is 80 g  $$mol^{-1}$$]

We need to find the percentage of bromine in an organic compound using the Carius method.

In the Carius method for halogen estimation, the organic compound is heated with fuming nitric acid in the presence of silver nitrate, converting halogens into silver halides. For bromine this reaction can be represented as $$\text{Organic-Br} \rightarrow AgBr$$.

We know the following data: mass of the organic compound is 0.25 g, mass of AgBr formed is 0.15 g, molar mass of Ag is 108 g/mol, molar mass of Br is 80 g/mol, and hence the molar mass of AgBr is 108 + 80 = 188 g/mol.

Therefore the moles of AgBr formed are calculated by $$\text{Moles of AgBr} = \frac{0.15}{188}$$. Since one mole of AgBr contains one mole of Br, the moles of bromine are also $$\text{Moles of Br} = \frac{0.15}{188}$$.

Substituting into the mass calculation gives $$\text{Mass of Br} = \frac{0.15}{188} \times 80 = \frac{12}{188} = \frac{12}{188} \approx 0.06383 \, \text{g}$$.

Next, the percentage of bromine in the compound is given by $$\% \text{Br} = \frac{\text{Mass of Br}}{\text{Mass of compound}} \times 100 = \frac{0.06383}{0.25} \times 100$$, which yields $$\% \text{Br} = 25.53\%$$.

Expressing this in the required format $$\_ \times 10^{-1}$$ gives $$25.53 = 255.3 \times 10^{-1}$$. Rounding to the nearest integer results in $$255 \times 10^{-1}$$.

The answer is 255.