In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr. The mass percentage of bromine in the compound is __ (Atomic mass, Ag = 108, Br = 80g mol^–1).

Given data:It is given to find the mass percentage of bromine in the compound by estimation of bromine by Carius method when 1.6 g of an organic compound gave 1.88 g of AgBr.

Explanation:

 1 mole of AgBr i.e 188 gmol-1 of AgBr has bromine content = 80 gmol-1
1 gmol-1 of AgBr will have bromine content = $\frac{80}{188}$gmol-1
1.88 gmol-1 of AgBr will have bromine content = $\frac{80}{188}\times 1.88$ = 0.8 gmol-1
Now, Percentage of bromine = 0.8 $\times \frac{1}{1.6}\times 100$ = 50 % 

Hence, the correct option is option (A) 50.