0.20 ~g of an organic compound gave 0.12 ~g of AgBr By using Carius method, the percentage of bromine in the compound will be

Step-by-step solution

Given, Mass of an organic compound =0.20 ~g Mass of AgBr =0.12 ~g Molecular mass of AgBr =188 ~g ~mol ⁻¹188 ~g of AgBr contains 80 ~g of bromine 0.12 ~g of AgBr will contain = 80 188 0.12=0.05 ~g of bromine Percentage of bromine = 0.05 0.20 100=25 %