Principle of Carius method

In Carius method, halogen present in the organic compound is converted into silver halide.

Here, bromine is converted into $\mathrm{AgBr}$.

Molar mass of $\mathrm{AgBr}$

$M(\mathrm{AgBr}) = M(\mathrm{Ag}) + M(\mathrm{Br}) = 108 + 80 = 188$

Mass fraction of bromine in $\mathrm{AgBr}$

In $188\text{ g}$ of $\mathrm{AgBr}$, mass of bromine $=80\text{ g}$.

Therefore, in $0.188\text{ g}$ of $\mathrm{AgBr}$, mass of bromine is

$\text{Mass of Br} = 0.188 \times \frac{80}{188}$

Since $0.188 = \frac{188}{1000}$,

$\text{Mass of Br} = \frac{188}{1000} \times \frac{80}{188} = \frac{80}{1000} = 0.08\text{ g}$

Percentage of bromine in the organic compound

Mass of organic compound taken $=0.2\text{ g}$.

$\%\text{Br} = \frac{0.08}{0.2} \times 100 = 40\%$

Nearest integer

$\boxed{40}$