Question

0.15 g of an organic compound gave 0.12g of AgBr by Carius method. The % of bromine in the Compound is(Molar mass of AgBr=188)

• A. 42.06
• B. 38.96
• C. 24.08
• D. 34.04

Answer 1

The correct option is D 34.04

$\%ofbromine=\frac{80\times massofAgBr\times 100}{188\times 0.15}=\frac{80\times 0.12\times 100}{188\times 0.15}=34.04$