Discussion Forum

In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is [ at. mass Ag=108, Br=80]

Answer:

Explanation:

Given,

  weight of organic compound =250  mg

 Weight of AgBr=141 mg

 $\therefore$   According to formula of %  of bromine by Carius method

       %  of Br=   Atomic weight of Br / Molecular  weight of AgBr

                 x    Weight of AgBr/ Weight of organic bromide   x 100

   $\therefore$   % of Br=    $\frac{80}{188}\times\frac{141}{250}\times 100$

                                    =  $\frac{1128000}{47000}=24$ %