In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is _________. (Nearest integer)
[Atomic mass: Ag = 108, Br = 80]

In Carius method, the halogen present in the organic compound is converted completely into its silver halide. Here the silver halide obtained is silver bromide, $$\mathrm{AgBr}$$.

The given data are:
    Mass of organic compound, $$w = 0.2 \text{ g}$$
    Mass of silver bromide, $$m_{\mathrm{AgBr}} = 0.188 \text{ g}$$
    Atomic mass of silver, $$M_{\mathrm{Ag}} = 108$$
    Atomic mass of bromine, $$M_{\mathrm{Br}} = 80$$

First, we need the molar mass of silver bromide. By definition,

$$M_{\mathrm{AgBr}} = M_{\mathrm{Ag}} + M_{\mathrm{Br}} = 108 + 80 = 188 \text{ g mol}^{-1}.$$

Now, to obtain the mass of bromine actually present in the precipitate, we use the proportion coming from the formula weight. The fraction of bromine in one mole of $$\mathrm{AgBr}$$ is

$$\frac{M_{\mathrm{Br}}}{M_{\mathrm{AgBr}}} = \frac{80}{188}.$$

Therefore, the mass of bromine in the given precipitate is

$$m_{\mathrm{Br}} = \frac{80}{188}\times m_{\mathrm{AgBr}}.$$

Substituting $$m_{\mathrm{AgBr}} = 0.188 \text{ g}$$, we get

$$m_{\mathrm{Br}} = \frac{80}{188}\times 0.188 = 0.188 \times \frac{80}{188} = 0.188 \times 0.4255319 = 0.08 \text{ g}.$$

We now calculate the percentage of bromine in the original compound sample. By definition,

$$\%\ \mathrm{Br} = \frac{\text{mass of Br in sample}}{\text{mass of sample}}\times 100 = \frac{0.08}{0.2}\times 100 = 0.4 \times 100 = 40\%.$$

After rounding to the nearest integer, the percentage remains $$40$$.

Hence, the correct answer is Option C.