In Carius method for estimation of halogens, 0.2g 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]

Given data:

           Mass of organic compound = 0.2 g
           Mass of AgBr = 0.188 g

To find = Percentage of bromine .

Explanation:

We know, Molar mass of AgBr = 188 g mol-1
In 1 mole of AgBr, bromine content = 80 g mol-1
In 1 g of AgBr, bromine content = $\frac{80}{188}$g mol-1
In 0.188 g of AgBr, bromine content = $\frac{80}{188}\times 0.188$ g mol-1
Percentage of bromine in the given compound = $\frac{80}{188}\times 0.188\times \frac{1}{0.2}\times 100 \% = 40 \% \left(approx\right)$

Hence, the correct option is option (D) 40.