In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg AgBr . What is the percentage of bromine in the compound (atomic mass of Ag = 108 and atomic number of Br = 80 )?

Step-by-step solution

R - Br → Carius method AgBr 250   mg organic compound is RBr . 141   mg AgBr ⇒ 141 × 80 188 mg Br From principle of atom coservation(POAC), mass of Bromine is conserved in reactant and product So, mass of Bromine in organic compound = 141 × 80 188   mg % of Br in organic compound = mass   of   Br mass   of   compound × 100 = 1 4 1 × 8 0 1 8 8 × 1 2 5 0 × 1 0 0 = 24%