devarshi-dt-logo

Question:

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: (Atomic mass Ag = 108; Br = 80)

24

36

48

60

Solution:

The molar mass of AgBr is 108 + 80 = 188 g/mol
Hence, 141 mg of AgBr will contain 141mg × (80g/188g) = 60 mg of bromine.
Hence, the percentage of bromine in the compound is (60mg/250mg) × 100 = 24